刷题79. Word Search

一、题目说明

题目79. Word Search，给定一个由字符组成的矩阵，从矩阵中查找一个字符串是否存在。可以连续横、纵找。不能重复使用，难度是Medium。

二、我的解答

惭愧，我写了很久总是有问题，就先看正确的写法，下面是回溯法的代码：

class Solution {
public:
	int m,n;
	//左 上 右 下
	int dx[4] = {-1,0,1,0};
	int dy[4] = {0,1,0,-1};
    bool exist(vector<vector<char>>& board,string word){
    	if(board.empty()||word.empty()){
    		return false;
		}
    	m = board.size();
    	n = board[0].size();
    	for(int i=0;i<m;i++){
    		for(int j=0;j<n;j++){
    			if(dfs(board,i,j,word,0)){
    				return true;
				}
			}
		}
		return false;
	}
	bool dfs(vector<vector<char>>& board,int x,int y,string&word,int pos){
		if(word[pos]!=board[x][y]){
			return false;
		}
		if(pos == word.size()-1){
			return true;
		}
		board[x][y] = '.';
		for(int i=0;i<4;i++){
			int nx = x + dx[i];
			int ny = y + dy[i];
			if(nx<m && nx>=0 && ny<n && ny>=0){
				if(dfs(board,nx,ny,word,pos+1)){
					return true;
				}
			}
		}
		board[x][y] = word[pos];
		return false;
	}
};

性能：

Runtime: 24 ms, faster than 87.44% of C++ online submissions for Word Search.
Memory Usage: 9.8 MB, less than 100.00% of C++ online submissions for Word Search.

三、优化措施

我的思路是用unordered_map<char,vector<vector<int>>> ump;来存储board中所有字符的出现位置，然后从word的第1个开始起查找，用dfs算法（回溯算法）进行匹配，修改并提交了差不多10次，才成功。

class Solution{
	public:
		bool exist(vector<vector<char>>& board,string word){

			if (board.empty() || word.empty()) {
	            return false;
	        }
			int row = board.size();
			int col = board[0].size();

	        if (row * col < word.length()) {
	            return false;
	        }

			for(int i=0;i<row;i++){
				for(int j=0;j<col;j++){
					char ch = board[i][j];
					ump[ch].push_back({i,j});
				}
			}

			if(dfs(board,0,0,0,word)){
				return true;
			}else{
				return false;
			}
		}
		bool dfs(vector<vector<char>>& board,int start,int x,int y,string& word){
			char ch = word[start];
			bool matched = false;
			if(ump.count(ch)){
				for(auto current: ump[ch]){
					int row1 = current[0];
					int col1 = current[1];
					//是否相邻
					if(start==0 || x==row1 && abs(y-col1)==1 || y==col1 && abs(x-row1)==1){
						if(board[row1][col1]!='.'){
							board[row1][col1] = '.';

							if(start<word.size()-1){
								matched = dfs(board,start+1,row1,col1,word);
								if(matched) return true;
							}else if(start==word.size()-1){
								return true;
							}

							board[row1][col1] = ch;
						}
					}
				};
			}else{
				return false;
			}

			return false;
		};
	private:
	    unordered_map<char,vector<vector<int>>> ump;
};

惭愧的是，性能还不如普通的回溯法：

Runtime: 764 ms, faster than 5.02% of C++ online submissions for Word Search.
Memory Usage: 169.2 MB, less than 16.18% of C++ online submissions for Word Search.