劍指 Offer 12. 矩陣中的路徑(中等)
阿新 • • 發佈:2021-08-30
通過率45.3%
題目描述:
給定一個m x n 二維字元網格board 和一個字串單詞word 。如果word 存在於網格中,返回 true ;否則,返回 false 。
單詞必須按照字母順序,通過相鄰的單元格內的字母構成,其中“相鄰”單元格是那些水平相鄰或垂直相鄰的單元格。同一個單元格內的字母不允許被重複使用。
例如,在下面的 3×4 的矩陣中包含單詞 "ABCCED"(單詞中的字母已標出)。
示例 1:
輸入:board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
輸出:true
示例 2:
輸入:board = [["a","b"],["c","d"]], word = "abcd"
輸出:false
提示:
1 <= board.length <= 200
1 <= board[i].length <= 200
board 和 word 僅由大小寫英文字母組成
思路:
深搜
1 /*JavaScript*/ 2 /** 3 * @param {character[][]} board 4 * @param {string} word 5 * @return {boolean} 6 */ 7 var dfs = function(board, n, m, i, j, word, k) {8 if(k === word.length) return true 9 if(i < 0 || j < 0 || i >= n || j >= m || board[i][j] !== word[k]) return false 10 board[i][j] = '*' 11 const res = dfs(board, n, m, i-1, j, word, k+1) || dfs(board, n, m, i, j+1, word, k+1) || dfs(board, n, m, i+1, j, word, k+1) || dfs(board, n, m, i, j-1, word, k+1)12 board[i][j] = word[k] 13 return res 14 } 15 16 var exist = function(board, word) { 17 const n = board.length 18 const m = board[0].length 19 for(let i = 0; i < n; i++) { 20 for(let j = 0; j < m; j++) { 21 if(dfs(board, n, m, i, j, word, 0)) return true 22 } 23 } 24 return false 25 };