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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
,"SEE"
,
-> returns true
,"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; } };
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原文地址:http://blog.csdn.net/u011345136/article/details/44937899