leetcode70word-search

题目描述

给出一个二维字符数组和一个单词，判断单词是否在数组中出现，

单词由相邻单元格的字母连接而成，相邻单元指的是上下左右相邻。同一单元格的字母不能多次使用。

例如：

给出的字符数组=

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

单词 ="ABCCED", -> 返回 true,
单词 ="SEE", ->返回 true,
单词 ="ABCB", -> 返回 false.

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", -> 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;
        
    }
};