yasinshaw
4247e0700d
Changes: - Removed all Java code implementations - Kept only Go language solutions - Renamed "## Go 解法" to "## 解法" - Removed "### Go 代码要点" sections - Cleaned up duplicate headers and empty sections - Streamlined documentation for better readability Updated files (9): - 三数之和.md - 两数相加.md - 无重复字符的最长子串.md - 最长回文子串.md - 括号生成.md - 子集.md - 单词搜索.md - 电话号码的字母组合.md - 柱状图中最大的矩形.md All 22 LeetCode Hot 100 Medium problems now use Go exclusively. Code is cleaner, more focused, and easier to follow. Generated with [Claude Code](https://claude.ai/code) via [Happy](https://happy.engineering) Co-Authored-By: Claude <noreply@anthropic.com> Co-Authored-By: Happy <yesreply@happy.engineering>
369 lines
7.8 KiB
Markdown
369 lines
7.8 KiB
Markdown
# 子集 (Subsets)
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## 题目描述
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给你一个整数数组 `nums`，数组中的元素 **互不相同**。返回该数组所有可能的子集（幂集）。
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解集 **不能** 包含重复的子集。你可以按 **任意顺序** 返回解集。
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### 示例
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**示例 1：**
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```
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输入：nums = [1,2,3]
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输出：[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
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```
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**示例 2：**
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```
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输入：nums = [0]
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输出：[[],[0]]
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```
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### 约束条件
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- `1 <= nums.length <= 10`
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- `-10 <= nums[i] <= 10`
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- `nums` 中的所有元素 **互不相同**
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## 解题思路
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### 方法一：回溯法（推荐）
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**核心思想：**对于每个元素，可以选择包含或不包含。使用回溯法生成所有可能的组合。
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**算法步骤：**
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1. 初始化结果数组和当前子集
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2. 定义回溯函数 `backtrack(start)`：
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- 将当前子集加入结果
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- 从 `start` 开始遍历，依次尝试包含每个元素
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- 递归调用后撤销选择（回溯）
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### 方法二：迭代法（位掩码）
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**核心思想：**子集可以用二进制表示。对于 n 个元素，共有 2^n 个子集。
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**算法步骤：**
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1. 计算子集总数 `total = 1 << n`
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2. 对于每个数字 `i` 从 0 到 `total-1`：
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- 将 `i` 的二进制表示转换为子集
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- 第 `j` 位为 1 表示包含 `nums[j]`
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### 方法三：级联法
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**核心思想：**对于已有的每个子集，通过添加当前元素生成新的子集。
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**算法步骤：**
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1. 初始化结果为 `[[]]`
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2. 对于每个元素：
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- 取出所有已有子集
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- 将当前元素添加到每个子集
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- 将新子集加入结果
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## 代码实现
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### Go 实现（回溯法）
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```go
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package main
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import "fmt"
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func subsets(nums []int) [][]int {
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result := [][]int{}
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current := []int{}
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var backtrack func(start int)
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backtrack = func(start int) {
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// 将当前子集加入结果（需要复制）
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temp := make([]int, len(current))
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copy(temp, current)
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result = append(result, temp)
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// 从 start 开始尝试包含每个元素
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for i := start; i < len(nums); i++ {
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// 选择当前元素
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current = append(current, nums[i])
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// 递归处理下一个元素
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backtrack(i + 1)
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// 撤销选择（回溯）
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current = current[:len(current)-1]
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}
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}
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backtrack(0)
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return result
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}
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// 测试用例
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func main() {
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// 测试用例1
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nums1 := []int{1, 2, 3}
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fmt.Printf("输入: %v\n", nums1)
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fmt.Printf("输出: %v\n", subsets(nums1))
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// 测试用例2
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nums2 := []int{0}
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fmt.Printf("\n输入: %v\n", nums2)
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fmt.Printf("输出: %v\n", subsets(nums2))
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// 测试用例3
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nums3 := []int{1, 2}
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fmt.Printf("\n输入: %v\n", nums3)
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fmt.Printf("输出: %v\n", subsets(nums3))
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}
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```
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```go
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func subsetsBitMask(nums []int) [][]int {
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n := len(nums)
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total := 1 << n // 2^n 个子集
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result := make([][]int, 0, total)
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for mask := 0; mask < total; mask++ {
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subset := []int{}
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for i := 0; i < n; i++ {
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// 检查第 i 位是否为 1
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if mask&(1<<i) != 0 {
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subset = append(subset, nums[i])
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}
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}
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result = append(result, subset)
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}
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return result
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}
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```
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```go
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func subsetsCascade(nums []int) [][]int {
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result := [][]int{{}} // 初始化为空集
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for _, num := range nums {
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// 对于每个已有子集，添加当前元素生成新子集
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newSubsets := make([][]int, 0, len(result))
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for _, subset := range result {
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newSubset := make([]int, len(subset)+1)
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copy(newSubset, subset)
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newSubset[len(subset)] = num
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newSubsets = append(newSubsets, newSubset)
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}
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result = append(result, newSubsets...)
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}
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return result
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}
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```
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## 复杂度分析
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### 回溯法
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- **时间复杂度：** O(n × 2^n)
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- 共有 2^n 个子集
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- 每个子集的复制需要 O(n) 时间
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- **空间复杂度：** O(n)
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- 递归栈深度最大为 n
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- 不包括存储结果的空间
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### 迭代法（位掩码）
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- **时间复杂度：** O(n × 2^n)
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- 需要生成 2^n 个子集
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- 每个子集需要 O(n) 时间构建
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- **空间复杂度：** O(1)
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- 只使用了常数级别的额外空间（不包括结果）
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### 级联法
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- **时间复杂度：** O(n × 2^n)
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- 每次迭代都会将子集数量翻倍
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- 总共需要处理 n 次
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- **空间复杂度：** O(n × 2^n)
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- 需要存储所有子集
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## 进阶问题
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### Q1: 如果数组中有重复元素，应该如何处理？
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**A:** 需要先排序，然后在回溯时跳过重复元素。
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```go
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func subsetsWithDup(nums []int) [][]int {
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sort.Ints(nums)
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result := [][]int{}
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current := []int{}
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var backtrack func(start int)
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backtrack = func(start int) {
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temp := make([]int, len(current))
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copy(temp, current)
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result = append(result, temp)
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for i := start; i < len(nums); i++ {
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// 跳过重复元素
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if i > start && nums[i] == nums[i-1] {
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continue
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}
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current = append(current, nums[i])
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backtrack(i + 1)
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current = current[:len(current)-1]
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}
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}
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backtrack(0)
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return result
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}
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```
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### Q2: 如果要求子集的大小恰好为 k，应该如何修改？
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**A:** 在回溯时添加终止条件。
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```go
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func subsetsK(nums []int, k int) [][]int {
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result := [][]int{}
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current := []int{}
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var backtrack func(start int)
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backtrack = func(start int) {
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if len(current) == k {
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temp := make([]int, len(current))
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copy(temp, current)
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result = append(result, temp)
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return
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}
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for i := start; i < len(nums); i++ {
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current = append(current, nums[i])
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backtrack(i + 1)
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current = current[:len(current)-1]
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}
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}
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backtrack(0)
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return result
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}
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```
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## P7 加分项
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### 1. 深度理解：为什么子集问题适合用回溯法？
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**回溯法的本质：**
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- 在解空间树中进行深度优先搜索
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- 每个节点代表一个决策（包含或不包含当前元素）
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- 通过撤销选择（回溯）来探索所有可能
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**为什么适合子集问题：**
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1. **决策清晰：**每个元素只有两种选择（包含或不包含）
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2. **无后效性：**当前选择不影响之前的选择
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3. **边界明确：**子集大小从 0 到 n
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### 2. 实战扩展：组合与排列
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**组合问题：**从 n 个元素中选 k 个，不考虑顺序
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**排列问题：**从 n 个元素中选 k 个，考虑顺序
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```go
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// 组合
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func combine(n int, k int) [][]int {
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result := [][]int{}
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current := []int{}
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var backtrack func(start int)
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backtrack = func(start int) {
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if len(current) == k {
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temp := make([]int, len(current))
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copy(temp, current)
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result = append(result, temp)
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return
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}
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for i := start; i <= n; i++ {
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current = append(current, i)
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backtrack(i + 1)
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current = current[:len(current)-1]
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}
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}
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backtrack(1)
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return result
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}
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// 排列
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func permute(nums []int) [][]int {
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result := [][]int{}
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current := []int{}
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used := make([]bool, len(nums))
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var backtrack func()
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backtrack = func() {
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if len(current) == len(nums) {
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temp := make([]int, len(current))
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copy(temp, current)
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result = append(result, temp)
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return
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}
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for i := 0; i < len(nums); i++ {
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if used[i] {
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continue
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}
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current = append(current, nums[i])
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used[i] = true
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backtrack()
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current = current[:len(current)-1]
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used[i] = false
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}
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}
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backtrack()
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return result
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}
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```
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### 3. 变形题目
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#### 变形1：子集 II（有重复元素）
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**LeetCode 90:** 给定一个可能包含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。
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```go
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func subsetsWithDup(nums []int) [][]int {
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sort.Ints(nums)
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result := [][]int{}
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current := []int{}
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var backtrack func(start int)
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backtrack = func(start int) {
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temp := make([]int, len(current))
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copy(temp, current)
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result = append(result, temp)
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for i := start; i < len(nums); i++ {
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if i > start && nums[i] == nums[i-1] {
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continue
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}
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current = append(current, nums[i])
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backtrack(i + 1)
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current = current[:len(current)-1]
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}
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}
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backtrack(0)
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return result
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}
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```
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### 4. 相关题目推荐
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- LeetCode 78: 子集（本题）
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- LeetCode 90: 子集 II
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- LeetCode 77: 组合
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- LeetCode 46: 全排列
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- LeetCode 47: 全排列 II
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