标签：style blog http io ar color os 使用 sp
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
Dynamic Programming String
注意项：
我写的如下：
1 #include <iostream> 2 #include <string> 3 #include <memory.h> 4 using namespace std; 5 6 class Solution { 7 public: 8 int minDistance(string word1, string word2) { 9 int len1 = word1.length(),len2=word2.length(); 10 if(len1==0) return len2; 11 if(len2==0) return len1; 12 int **dpmap = new int *[len1+1]; 13 dpmap[0] =new int[(len1+1)*(len2+1)]; 14 memset(dpmap[0],0,sizeof(int)*(len1+1)*(len2+1)); 15 for(int i= 1;i<=len1;i++) 16 dpmap[i] = dpmap[i-1]+len2+1; 17 for(int i=0;i<=len1;i++) 18 dpmap[i][0] = i; 19 for(int j=0;j<=len2;j++) 20 dpmap[0][j] = j; 21 for(int i=1;i<=len1;i++){ 22 for(int j=1;j<=len2;j++){ 23 if(word1[i-1]==word2[j-1]) dpmap[i][j]=dpmap[i-1][j-1]; 24 else{ 25 dpmap[i][j]=(dpmap[i-1][j]>dpmap[i][j-1]?dpmap[i][j-1]:dpmap[i-1][j])+1; 26 if(dpmap[i-1][j-1]+1<dpmap[i][j]) 27 dpmap[i][j] = dpmap[i-1][j-1]+1; 28 } 29 } 30 } 31 int ret = dpmap[len1][len2]; 32 // for(int i=0;i<=len1;i++){ 33 // for(int j=0;j<=len2;j++) 34 // cout<<dpmap[i][j]<<" "; 35 // cout<<endl; 36 // } 37 delete []dpmap[0]; 38 delete []dpmap; 39 return ret; 40 } 41 }; 42 43 int main() 44 { 45 string word1 = "123"; 46 string word2 = "111"; 47 Solution sol; 48 cout<<sol.minDistance(word1,word2)<<endl; 49 return 0; 50 }
标签：style blog http io ar color os 使用 sp
原文地址：http://www.cnblogs.com/Azhu/p/4128242.html
