leetcode79.单词搜索

题目

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。

用例

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

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

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

求解

/**
 * @param {character[][]} board
 * @param {string} word
 * @return {boolean}
 */
var exist = function(board, word) {
    let x_length = board.length
    let y_length = board[0].length
    let visited = new Array(x_length).fill(0)
    for(let i=0;i=0&&board[x-1][y]==word[index]&&visited[x-1][y]==0){
                visited[x-1][y]=1
                find_next_word(index+1,x-1,y)
                visited[x-1][y]=0
            }
            if(y-1>=0&&board[x][y-1]==word[index]&&visited[x][y-1]==0){
                visited[x][y-1]=1
                find_next_word(index+1,x,y-1)
                visited[x][y-1]=0
            }
            if(x+1