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Javascript(JS) leetcode 79. 单词搜索

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)

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