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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。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","d"]], word = "abcd"输出:false
class Solution { public boolean exist(char[][] board, String word) { //DFS+剪枝 char[] words=word.toCharArray(); int n=board.length; int m=board[0].length; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if(dfs(board,words,i,j,0)) return true; } } return false; } private static boolean dfs(char[][] board, char[] word, int i, int j, int k) { if(i>=board.length||i<0||j>=board[0].length||j<0||board[i][j]!=word[k]) return false; if(k==word.length-1) return true; //如果匹配的话 board[i][j]='\0'; boolean res=dfs(board,word,i+1,j,k+1)||dfs(board,word,i,j+1,k+1) ||dfs(board,word,i-1,j,k+1)||dfs(board,word,i,j-1,k+1); //开始回溯的时候恢复原样 board[i][j]=word[k]; return res; }}
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