*Leetcode 72 编辑距离

Posted 2023-02-03 Z-Pilgrim

力扣

dp[i][0] 和dp[j][0]需要单独处理

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <map>  
#include <set>  
#include <string>

using namespace std;

//class Solution 
//public:
//	int minDistance(string word1, string word2) 
//		// dp[i][j] : ->i ->j对齐需要的最小操作数
//		// dp[0][0]特殊处理
//		// w[i] == w[j]:
//		//     dp[i][j] = min(dp[i-1][j]+1, dp[i-1][j-1], dp[i][j-1]+1);
//		// w[i] != w[j]:
//		//     dp[i][j] = min(dp[i-1][j]+1, dp[i-1][j-1]+1, dp[i][j-1]+1);
//		if(word1.size() == 0) return word2.size();
//		if(word2.size() == 0) return word1.size();
//		int dp[word1.size()][word2.size()];
//		for (int i = 0; i < word1.size(); i++) 
//			for( int j = 0; j < word2.size(); j++) 
//				if(!i && !j) 
//					dp[i][j] = word1[i] == word2[j] ? 0:1;
//				 else 
//					dp[i][j] = 200000000;
//					if(i) dp[i][j] = min(dp[i][j], dp[i-1][j]+1);
//					if(j) dp[i][j] = min(dp[i][j], dp[i][j-1]+1);
//					if(i && j)
//						if(word1[i] == word2[j] ) 
//							dp[i][j] = min(dp[i][j], dp[i-1][j-1]);
//						 else 
//							 dp[i][j] = min(dp[i][j], dp[i-1][j-1]+1);
//						
//					 
//					printf("dp[%d][%d]=%d\\n", i+1, j+1, dp[i][j]);
//				
//			
//		 
//		return dp[word1.size()-1][word2.size()-1];
//	
//;

class Solution 
public:
	int minDistance(string word1, string word2) 
		// dp[i][j] : ->i ->j对齐需要的最小操作数
		// dp[0][0]特殊处理
		// w[i] == w[j]:
		//     dp[i][j] = min(dp[i-1][j]+1, dp[i-1][j-1], dp[i][j-1]+1);
		// w[i] != w[j]:
		//     dp[i][j] = min(dp[i-1][j]+1, dp[i-1][j-1]+1, dp[i][j-1]+1);
		if(word1.size() == 0) return word2.size();
		if(word2.size() == 0) return word1.size();
		int dp[word1.size()][word2.size()];
		for (int i = 0; i < word1.size(); i++) 
			for( int j = 0; j < word2.size(); j++) 
				if(!i && !j) 
					dp[i][j] = word1[i] == word2[j] ? 0:1;
				 else 
					dp[i][j] = 200000000;
					if(i) dp[i][j] = min(dp[i][j], dp[i-1][j]+1);
					if(j) dp[i][j] = min(dp[i][j], dp[i][j-1]+1);
					int v ;
					if(i && j)
						v = dp[i-1][j-1];
					 else 
						if(i) v = i;
						else v = j;
					
					if(word1[i] == word2[j] ) 
						dp[i][j] = min(dp[i][j], v);
					 else 
						dp[i][j] = min(dp[i][j], v+1);
					
					 printf("dp[%d][%d]=%d\\n", i+1, j+1, dp[i][j]);
				
			
		 
		return dp[word1.size()-1][word2.size()-1];
	
;

class Solution2 
public:
	int minDistance(string word1, string word2) 
		int n = word1.length();
		int m = word2.length();

		// 有一个字符串为空串
		if (n * m == 0) return n + m;

		// DP 数组
		vector<vector<int> > D(n + 1, vector<int>(m + 1));

		// 边界状态初始化
		for (int i = 0; i < n + 1; i++) 
			D[i][0] = i;
		
		for (int j = 0; j < m + 1; j++) 
			D[0][j] = j;
		

		// 计算所有 DP 值
		for (int i = 1; i < n + 1; i++) 
			for (int j = 1; j < m + 1; j++) 
				int left = D[i - 1][j] + 1;
				int down = D[i][j - 1] + 1;
				int left_down = D[i - 1][j - 1];
				if (word1[i - 1] != word2[j - 1]) left_down += 1;
				D[i][j] = min(left, min(down, left_down));
				printf("D[%d][%d]=%d\\n", i, j, D[i][j]);


			
		
		return D[n][m];
	
; 

int main(int argc, char *argv[]) 
	string a = "sea",b ="eat";
	Solution s1 = Solution();
	Solution2 s2 = Solution2();
	s1.minDistance(a,b);
	cout << "*******\\n\\n\\n*******" << endl;
	s2.minDistance(a,b);

	return 0;