leetcode 79. 单词搜索

jay神2023-10-16 13:43

2023.10.16

本题使用回溯法解决，回溯开头先判断i、j是否越界 and 字符是否不匹配。是的话则直接返回false。否则，开始回溯判断，后面的字符是否都相等，为了防止矩阵的字符重复使用，回溯前，需要将原来的字符用'#'代替，回溯完再换回来。 java代码如下：

java 复制代码

class Solution {
    private boolean backtrating(char[][] board,char[] word,int i,int j,int k){
        if(i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] != word[k]) return false;//越界或者字母不匹配
        if(k == word.length-1) return true;
        //两字母匹配 开始回溯
        char temp = board[i][j];
        board[i][j] = '#';
        boolean ans = backtrating(board,word,i+1,j,k+1) || backtrating(board,word,i-1,j,k+1) ||
                      backtrating(board,word,i,j+1,k+1) || backtrating(board,word,i,j-1,k+1);
        board[i][j] = temp;
        return ans;
    }

    public boolean exist(char[][] board, String word) {
        char[] words = word.toCharArray();
        for(int i=0; i<board.length; i++){
            for(int j=0; j<board[0].length; j++){
                if(backtrating(board,words,i,j,0)) return true;
            }
        }
        return false;
    }
}