yasinshaw
15dbd75004
- Remove all "## Java 解法" sections and Java code blocks - Replace "## Go 解法" with "## 解法" - Remove "### Go 代码要点" and "### Java 代码要点" sections - Keep all Go code sections intact - Maintain complete documentation structure and content - Update 22 markdown files in the LeetCode Hot 100 directory
133 lines
2.8 KiB
Markdown
133 lines
2.8 KiB
Markdown
# 单词搜索 (Word Search)
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## 题目描述
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给定一个 `m x n` 二维字符网格 `board` 和一个字符串单词 `word`。如果 `word` 存在于网格中，返回 `true`；否则，返回 `false`。
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单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中"相邻"单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
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### 示例
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**示例 1：**
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```
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输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
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输出：true
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```
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**示例 2：**
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```
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输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
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输出：true
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```
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**示例 3：**
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```
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输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
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输出：false
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```
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## 解题思路
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### 方法一：DFS + 回溯（推荐）
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**核心思想：**对每个位置进行 DFS，搜索是否存在匹配的单词路径。
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**算法步骤：**
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1. 遍历网格的每个位置
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2. 如果当前位置字符匹配单词首字符，开始 DFS
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3. DFS 过程中：
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- 标记当前已访问
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- 向四个方向递归搜索
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- 如果找到完整单词，返回 true
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- 回溯时撤销访问标记
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## 代码实现
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### Go 实现
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```go
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package main
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func exist(board [][]byte, word string) bool {
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m, n := len(board), len(board[0])
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visited := make([][]bool, m)
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for i := range visited {
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visited[i] = make([]bool, n)
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}
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var dfs func(i, j, k int) bool
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dfs = func(i, j, k int) bool {
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// 找到完整单词
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if k == len(word) {
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return true
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}
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// 边界检查或不匹配
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if i < 0 || i >= m || j < 0 || j >= n ||
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visited[i][j] || board[i][j] != word[k] {
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return false
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}
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// 标记访问
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visited[i][j] = true
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// 向四个方向搜索
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found := dfs(i+1, j, k+1) ||
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dfs(i-1, j, k+1) ||
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dfs(i, j+1, k+1) ||
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dfs(i, j-1, k+1)
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// 回溯：取消标记
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visited[i][j] = false
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return found
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}
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for i := 0; i < m; i++ {
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for j := 0; j < n; j++ {
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if board[i][j] == word[0] && dfs(i, j, 0) {
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return true
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}
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}
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}
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return false
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}
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```
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### Java 实现
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## 复杂度分析
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- **时间复杂度：** O(m × n × 4^L)
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- m × n 是网格大小
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- L 是单词长度
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- 最坏情况每个位置都要搜索 4 个方向
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- **空间复杂度：** O(L)
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- 递归栈深度最大为 L
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- visited 数组 O(m × n)
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## P7 加分项
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### 变形题目：单词搜索 II
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**LeetCode 212:** 给定一个 m x n 二维字符网格 board 和一个单词列表 words，返回所有在二维网格和字典中出现的单词。
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```go
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func findWords(board [][]byte, words []string) []string {
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// 构建 Trie 树
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trie := buildTrie(words)
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result := []string{}
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for i := 0; i < len(board); i++ {
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for j := 0; j < len(board[0]); j++ {
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dfsBoard(board, i, j, trie, &result)
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}
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}
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return result
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}
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```
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