79_WordSearch - a920604a/leetcode GitHub Wiki
title: 79. Word Search
終止條件 if(word.size() == k) return true;
class Solution {
public:
bool backtracking(vector<vector<char>>& board, string word, int i, int j,int k, vector<vector<bool>> & visited){
if(word.size() == k) return true;
int n = board.size() , m= board[0].size();
if(i<0 || j< 0 || i>n-1 || j>m-1 || visited[i][j]) return false;
if(board[i][j] !=word[k]) return false;
visited[i][j] = true;
bool ret = backtracking(board, word, i+1,j, k+1, visited) || \
backtracking(board, word, i,j+1, k+1, visited) || \
backtracking(board, word, i-1,j, k+1, visited) || \
backtracking(board, word, i,j-1, k+1, visited) ;
visited[i][j] = false;
return ret;
}
bool exist(vector<vector<char>>& board, string word) {
int n = board.size() , m= board[0].size();
vector<vector<bool>> visited(n,vector<bool>(m, false));
for(int i=0;i<n;++i){
for(int j=0;j<m;++j){
if(board[i][j] == word[0] && backtracking(board, word, i, j, 0, visited)) return true;
}
}
return false;
}
};
- 也可以不用visited 額外空間,只需要拜訪到的對board進行修改,結束後再恢復之前的狀態。
- time complexity
O(4*n^2)
- space complexity
O(n^2)