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【leetcode】Edit Distance
Edit Distance
Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character
采用动态规划求解
1. d[0, j] = j;
2. d[i, 0] = i;
3. d[i, j] = d[i-1, j - 1] if A[i] == B[j]
4. d[i, j] = min(d[i-1, j - 1], d[i, j - 1], d[i-1, j]) + 1 if A[i] != B[j]
1 class Solution { 2 public: 3 int minDistance(string word1, string word2) { 4 5 int n1=word1.length(); 6 int n2=word2.length(); 7 8 if(n1==0) return n2; 9 if(n2==0) return n1;10 11 //采用二维数组效率更高 12 //vector<vector<int> > dp(n1+1,vector<int>(n2+1));13 14 int **dp=new int*[n1+1];15 for(int i=0;i<n1+1;i++)16 {17 dp[i]=new int[n2+1];18 }19 20 21 dp[0][0]=0;22 for(int i=1;i<=n1;i++)23 {24 dp[i][0]=i;25 }26 for(int j=1;j<=n2;j++)27 {28 dp[0][j]=j;29 }30 31 for(int i=1;i<=n1;i++)32 {33 for(int j=1;j<=n2;j++)34 {35 36 if(word1[i-1]==word2[j-1])37 {38 dp[i][j]=dp[i-1][j-1];39 }40 else41 {42 dp[i][j]=min(dp[i-1][j],min(dp[i][j-1],dp[i-1][j-1]))+1;43 }44 }45 }46 47 int result=dp[n1][n2];48 for(int i=0;i<n1+1;i++)49 {50 delete[] dp[i];51 }52 53 return result;54 }55 };
【leetcode】Edit Distance
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