码迷,mamicode.com
首页 > 其他好文 > 详细

LeetCode Submission Details

时间:2015-04-08 13:20:30      阅读:124      评论:0      收藏:0      [点我收藏+]

标签:

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 =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

题意:在字符矩阵中可以走4个方向,查找一个单词是否存在。

思路:深搜+回溯。

class Solution {
public:
    int vis[1000][1000];
    int dx[4]={0, 0, 1, -1};
    int dy[4]={1, -1, 0, 0};

    bool check(vector<vector<char> > &a, string &w, int x, int y, int cur) {
        if (cur == w.size()) return true;
        if (x >= a.size() || x < 0 || y >= a[0].size() || y < 0) return false;
        if (vis[x][y] == 1) return false;
        if (w[cur] != a[x][y]) return false;
        vis[x][y] = 1;

        for (int i = 0; i < 4; i++) {
            if (check(a, w, x+dx[i], y+dy[i], cur+1)) return true;
        }
        vis[x][y] = 0;
        return false;
    }

    bool exist(vector<vector<char> > &board, string word) {
        int n = board.size();
        int m = board[0].size();
        if (n == 0 || m == 0) return false;
        for (int i = 0; i < n; i++) 
            for (int j = 0; j < m; j++)
                vis[i][j] = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (check(board, word, i, j, 0)) return true;
        return false;
    }
};



LeetCode Submission Details

标签:

原文地址:http://blog.csdn.net/u011345136/article/details/44937899

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!