leetcode70word-search
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题目描述
[↵ ["ABCE"],↵ ["SFCS"],↵ ["ADEE"]↵]单词 ="ABCCED", -> 返回 true,
单词 ="SEE", ->返回 true,
单词 ="ABCB", -> 返回 false.
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", -> returnstrue,
word ="SEE", -> returnstrue,
class Solution {
public:
bool isOut(int r,int c,int rows,int cols){
return c<0 || c>=cols || r<0 || r>=rows;
}
bool DFS(vector< vector< char >>&board,int r, int c,string &word,int start){
if (start>=word.size())
return true;
if (isOut(r,c,board.size(),board[0].size() )|| word[start]!=board[r][c])
return false;
int dx[]={0,0,1,-1},dy[]={1,-1,0,0};
char tmp=board[r][c];
board [r][c]=‘.‘;
for (int i=0;i<4;++i){
if (DFS(board,r+dx[i],c+dy[i],word,start+1))
return true;
}
board[r][c]=tmp;
return false;
}
bool exist(vector<vector<char> > &board, string word) {
int rows=board.size(),cols=board[0].size();
for (int r=0;r<rows;++r)
for (int c=0;c<cols;++c){
if (board[r][c]==word[0])
if (DFS(board,r,c,word,0))
return true;
}
return false;
}
};
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