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最新推荐文章于 2018-07-25 01:06:22 发布

Joyyiwei

最新推荐文章于 2018-07-25 01:06:22 发布

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分类专栏： LeetCode与面试题

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本文介绍了一个二维网格中查找单词的问题及解决方案。通过深度优先搜索(DFS)与回溯的方法，在字符矩阵中寻找指定的单词路径，确保每个字母仅被使用一次。

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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;
    }
};