leetcode79.单词搜索

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<x_length;i++){
        visited[i] = new Array(y_length).fill(0)
    }
    let res = false
    //参数为
    //寻找第几个字母
    //上一个字母的横纵坐标
    function find_next_word(index,x,y){
        if(index==word.length){
            res = true
        }
        if(index==0){
            //寻找第一个字母
            for(let i=0;i<x_length;i++){
                for(let j=0;j<y_length;j++){
                    if(board[i][j]==word[index]){
                        visited[i][j]=1
                        find_next_word(index+1,i,j)
                        visited[i][j]=0
                    }
                }
            }
        }else{
            //寻找下一个字母
            if(x-1>=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<x_length&&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<y_length&&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
            }
        }
    }
    find_next_word(0,-1,-1)
    return res
};