LeetCode 79

Posted 2021-02-21 菜鸡的世界

https://leetcode-cn.com/problems/word-search/

这个题没啥好说的，从每个字母开始去DFS看是不是可以找得到同样内容的字符串，但是在实现的细节上还是有很多需要改进的地方。

首先先上自己的代码

class Solution {
    public boolean exist(char[][] board, String word) {
        if(board == null || board.length == 0 || word == null || word.length() == 0){
            return false;
        }
        int[][] direction = new int[][]{{-1,0},{1,0},{0,-1},{0,1}};
        boolean[][] visited = new boolean[board.length][board[0].length];
        for(int i = 0;i < board.length;i++){
            for(int j = 0; j < board[i].length;j++){
                if(travel(board, word, 0, visited, direction, i, j)){
                    return true;
                }
            }
        }
        return false;
    }
    private boolean travel(char[][] board, String word, int pos, boolean[][] visited, int[][] direction,int i , int j){
        if(i<0||i>=board.length || j<0 || j >= board[0].length || visited[i][j] || board[i][j] != word.charAt(pos)){
            return false;
        }
        if(pos == word.length()-1){
            return true;
        }
        visited[i][j] = true;
        boolean flag = false;
        for(int k = 0; k< 4;k++){
            int newX = i+direction[k][0];
            int newY = j+direction[k][1];
            flag = travel(board, word, pos + 1, visited, direction, newX, newY) || flag;
        }
        if(!flag){
            visited[i][j] = false;
        }
        return flag;
    }
}

View Code

可以看到我用到visited数组来记录是否访问过。

再贴其他人的代码

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

    boolean search(char[][] board, String word, int i, int j, int k) {
        if (k >= word.length()) return true;
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] != word.charAt(k)) return false;
        board[i][j] += 256;
        boolean result = search(board, word, i - 1, j, k + 1) || search(board, word, i + 1, j, k + 1)
                || search(board, word, i, j - 1, k + 1) || search(board, word, i, j + 1, k + 1);
        board[i][j] -= 256;
        return result;
    }
}

View Code

他的代码里去掉了visited数组，减少了内存的开销，加快程序的运行速度，太顶了这个复用board数组的方法！！！！