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 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
var exist = function(board, word) {
function dfs(x,y,s){
if(s === word.length) return true
if(x < 0 || x >= board.length || y < 0 || y >= board[0].length) return false
if(board[x][y] !== word[s]) return false
let char = board[x][y]
board[x][y] = '*'
let res = dfs(x-1,y,s+1) || dfs(x+1,y,s+1) || dfs(x,y-1,s+1) || dfs(x,y+1,s+1)
board[x][y] = char
return res
}
for(let i =0;i < board.length;i++){
for(let j =0;j < board[0].length;j++){
if (board[i][j] === word[0]) {
if(dfs(i, j, 0)){
return true
}
}
}
}
return false
};
时间复杂度:O(MN3^L)),其中 M,,N 为网格的长度与宽度,L 为字符串 word 的长度。
空间复杂度:O(1)