79. 单词搜索 (opens new window)

Description

Difficulty: 中等

Related Topics: 数组 (opens new window), 回溯 (opens new window), 矩阵 (opens new window)

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 。

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。

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true

示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board 和 word 仅由大小写英文字母组成

**进阶：**你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？

Solution

Language: JavaScript

/**
 * @param {character[][]} board
 * @param {string} word
 * @return {boolean}
 */
// 记录该元素已使用
// 上下左右不能超边界
// 如果没找到最终指的时候恢复上一个节点的值
var exist = function(board, word) {
 // 越界处理
 board[-1] = [] // 这里处理比较巧妙，利用了js的特性
 board.push([])

 // 寻找首个字母
 for (let y = 0; y < board.length; y++) {
  for (let x = 0; x < board[0].length; x++) {
   if (word[0] === board[y][x] && dfs(word, board, y, x, 0)) return true
  }
 }
 return false
};

const dfs = function(word, board, y, x, i) {
 if (i + 1 === word.length) return true
 var tmp = board[y][x]
 // 标记该元素已使用
 board[y][x] = false
 if (board[y][x+1] === word[i+1] && dfs(word, board, y, x+1, i+1)) return true
 if (board[y][x-1] === word[i+1] && dfs(word, board, y, x-1, i+1)) return true
 if (board[y+1][x] === word[i+1] && dfs(word, board, y+1, x, i+1)) return true
 if (board[y-1][x] === word[i+1] && dfs(word, board, y-1, x, i+1)) return true
 // 回溯
 board[y][x] = tmp
}