/***** 定义状态: DP[i][j]其中i表示word1前i个字符,j表示Word2前i个字符 DP[i][j]表示单词1前i个字符匹配单词2前j个字符,最少变换次数; 状态转移: for i:[0,m] for j:[0,n] if(word1[i-1]==word2[j-1]) DP[i][j]=DP[i-1][j-1]; else DP[i][j]=min(DP[i-1][j],DP[i][j-1],DP[i-1][j-1])+1; return DP[m][n]; ******/ class Solution { public: int minDistance(string word1, string word2) { int m=word1.size(),n=word2.size(); vector<vector<int> > DP(m+1,vector(n+1,0)); //初始化 for(int i=0;i<=m;i++){ DP[i][0]=i; } for(int j=0;j<=n;j++){ DP[0][j]=j; } //状态转移 for(int i=1;i<=m;i++){ for(int j=1;j<=n;j++){ if(word1[i-1]==word2[j-1]) DP[i][j]=DP[i-1][j-1]; else DP[i][j]=min(min(DP[i-1][j],DP[i][j-1]),DP[i-1][j-1])+1; } } return DP[m][n]; } };