【LeetCode】Word Search

参考链接

http://blog.csdn.net/doc_sgl/article/details/13008175

题目描述

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 =

[  ["ABCE"],  ["SFCS"],  ["ADEE"]]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

题目分析

总结

代码示例

class Solution {public:    bool exist(vector<vector<char> > &board, string word) {        int m = board.size();        if(m < 1) return false;        if(word.size() < 1) return false;        int n = board[0].size();        vector<vector<bool> > used(m, vector<bool>(n,false));        for(int i = 0;i<m;i++)        {            for(int j = 0;j<n;j++)            {                if(board[i][j] == word[0] && existCore(board,word,i,j,0,used))                    return true;            }        }        return false;    }    bool existCore(vector<vector<char> > &board, string word,int i,int j,int index,vector<vector<bool> > &used)    {        if(index == word.size())///////////////            return true;                if(i<0 || i>=board.size() || j <0 || j>=board[0].size())            return false;                if(used[i][j] || board[i][j] != word[index])            return false;                used[i][j] = true;        if(existCore(board,word,i+1,j,index+1,used)) return true;/////        if(existCore(board,word,i-1,j,index+1,used)) return true;//////        if(existCore(board,word,i,j+1,index+1,used)) return true;///////        if(existCore(board,word,i,j-1,index+1,used)) return true;//////////        used[i][j] = false;        return false;    }};

bool dfs(vector<vector<char> > &board, int x, int y, string word){if(word.size() == 0)return true;bool flag = false;if(x-1>=0 && board[x-1][y] == word[0]){board[x-1][y] = '#';flag = dfs(board, x-1, y, word.substr(1));board[x-1][y] = word[0];}if(!flag && y-1>=0 && board[x][y-1] == word[0]){board[x][y-1] = '#';flag = dfs(board, x, y-1, word.substr(1));board[x][y-1] = word[0];}if(!flag && x+1<board.size() && board[x+1][y] == word[0]){board[x+1][y] = '#';flag = dfs(board, x+1, y, word.substr(1));board[x+1][y] = word[0];}if(!flag && y+1<board[0].size() && board[x][y+1] == word[0]){board[x][y+1] = '#';flag = dfs(board, x, y+1, word.substr(1));board[x][y+1] = word[0];}return flag;}bool exist(vector<vector<char> > &board, string word) {        // Note: The Solution object is instantiated only once.        if(word.size() < 1)return true;int row = board.size();int col = board[0].size();set<string> st;for(int i = 0; i < row; i++)for(int j = 0; j < col; j++)if(board[i][j] == word[0]){board[i][j] = '#';if(dfs(board,i,j,word.substr(1)))return true;board[i][j] = word[0];}return false;    }

推荐学习C++的资料

C++标准函数库

http://download.csdn.net/detail/chinasnowwolf/7108919

在线C++API查询

http://www.cplusplus.com/