79. 单词搜索
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I am ne zha / Jeskson commented
79. 单词搜索
Description
Difficulty: 中等
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
// 如果没找到最终值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
}