Word Search leetcode

lpy1990

Word Search

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.

回溯法

说明

从每一个点出发, 上下左右搜索看是否能找到相等于word的字符串. 因为每一次的访问的点不能被二次访问, 所以要维护一个m * n 的boolean矩阵记录访问过的点.

复杂度

因为对于每个点来说复杂度都是O(m*n)
所以对于总体来说 时间O(m^2n^2) 空间是boolean数组复杂度O(m*n)

代码

public boolean exist(char[][] board, String word) {
    if (word == null || word.length() == 0) {
        return true;
    }
    if (board == null || board.length == 0 || board[0].length == 0) {
        return false;
    }
    boolean[][] visit = new boolean[board.length][board[0].length];
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
           if ( helper(board, visit, word, i, j, 0)) {
               return true;
           }
        }
    }
    return false;
}
public boolean helper(char[][] board, boolean[][] visit, String word, int i, int j, int pos) {
    if (pos == word.length()) {
        return true;
    }
    if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || visit[i][j] || board[i][j] != word.charAt(pos)) {
        return false;
    }
    visit[i][j] = true;
    boolean res = helper(board, visit, word, i + 1, j, pos + 1) || helper(board, visit, word, i - 1, j, pos + 1) || helper(board, visit, word, i , j - 1, pos + 1) || helper(board, visit, word, i, j + 1, pos + 1);
    visit[i][j] = false;
    return res;
}
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