79. 单词搜索
79. 单词搜索
给定一个 m x n
二维字符网格 board
和一个字符串单词 word
。如果 word
存在于网格中,返回 true
;否则,返回 false
。
单词必须按照字母顺序,通过相邻的单元格内的字母构成,其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
示例 1:
输入:board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出:true
示例 2:
输入:board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出:true
示例 3:
输入:board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出:false
提示:
m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board
和word
仅由大小写英文字母组成
进阶:你可以使用搜索剪枝的技术来优化解决方案,使其在 board
更大的情况下可以更快解决问题?
思路:
简单回溯算法。需要开辟一个数组记录是否走过某个点
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
vector<vector<bool>>used(board.size(),vector<bool>(board[0].size(),false));
int m=board.size();
int n=board[0].size();
bool flag=false;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(board[i][j]==word[0]){
flag=backtrack(board,i,j,used,word,0);
}
if(flag)return true;
}
}
return false;
}
bool backtrack(vector<vector<char>>& board,int x,int y,vector<vector<bool>>& used, string word,int index){
if(x<0||x>=board.size()||y<0||y>=board[0].size()||used[x][y]==true)return false;
if(board[x][y]!=word[index])return false;
if(index==word.size()-1)return true;
bool f=false;
//做选择
used[x][y]=true;
f=backtrack(board,x+1,y,used,word,index+1)||backtrack(board,x-1,y,used,word,index+1)||backtrack(board,x,y+1,used,word,index+1)||backtrack(board,x,y-1,used,word,index+1);//只要有一个成功即可
//撤销选择
used[x][y]=false;
return f;
}
};
本文来自博客园,作者:{BailanZ},转载请注明原文链接:https://www.cnblogs.com/BailanZ/p/16304489.html