leetcode 79. Word Search | Java最短代码实现

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word  =  "ABCCED" , -> returns  true ,
word  =  "SEE" , -> returns  true ,

word = "ABCB", -> returns false.

【抛砖】

必然要对每一点的每一条路径进行深度遍历，遍历过程中一旦出现：

1.数组越界、2.该点已访问过、3.该点的字符和word对应的index字符不匹配

就要对该路径进行剪枝：

    int[] dh = {0, 1, 0, -1};  //检索方向[右,下,左,上]
    int[] dw = {1, 0, -1, 0};
    public boolean exist(char[][] board, String word) {
        boolean[][] isVisited = new boolean[board.length][board[0].length];  //访问标记
        for (int i = 0; i < board.length; i++)
            for (int j = 0; j < board[0].length; j++)
                if (isThisWay(board, word, i, j, 0, isVisited)) return true;
        return false;
    }
    public boolean isThisWay(char[][] board, String word, int row, int column, int index, boolean[][] isVisited) {
        if (row < 0 || row >= board.length || column < 0 || column >= board[0].length
            || isVisited[row][column] || board[row][column] != word.charAt(index))
                return false;  //剪枝
        if (++index == word.length()) return true;  //word所有字符均匹配上
        isVisited[row][column] = true;
        for (int i = 0; i < 4; i++)
            if (isThisWay(board, word, row + dh[i], column + dw[i], index, isVisited))
                return true;  //以board[row][column]为起点找到匹配上word路径
        isVisited[row][column] = false;  //遍历过后，将该点还原为未访问过
        return false;
    }

87 / 87  test cases passed. Runtime: 11 ms  Your runtime beats 75.60% of javasubmissions.

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