leetcode79 word serach 解题报告

问题描述

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 =

[

['A','B','C','E'],

['S','F','C','S'],

['A','D','E','E']

]

word = "ABCCED", -> returns true,

word = "SEE", -> returns true,

word = "ABCB", -> returns false.

即在一个给定的字符的二维矩阵中寻找是否有给定的单词。每个字母必须相连，只能出现一次。

思路

对于给定字符串，在二维矩阵中进行深度优先搜索，不过要注意边界的判断。

代码

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

因为要求每个字母只能出现一次，所以设置了一个visited数组，来对字母是否出现过进行保存。如果在四个方向都判断是错误的，则设置visited为否。