LeetCode:Edit Distance（字符串编辑距离DP）

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

思路：题意求字符串的最小编辑距离。设状态为f[i][j],表示A[0][i] B[0][j]之间的最小编辑距离。

forexamplr:str1c str2d

1、c==d f[i][j]=f[i-1][j-1]2、c!=d　 (1)如果将c换成d 则f[i][j]=f[i-1][j-1]+1   (2)如果在c后面添加一个d f[i][j]=f[i][j-1]+1   (3)如果将c删除 f[i][j]=f[i-1][j-1]+1

简单的状态方程为

dp[i, j] = min { dp[i - 1, j] + 1,  dp[i, j - 1] + 1,  dp[i - 1, j - 1] + (s[i] == t[j] ? 0 : 1) }

 1 class Solution {
 2 public:
 3     int minDistance(string word1, string word2) {
 4
 5         const int n=word1.size();
 6         const int m=word2.size();
 7
 8         vector<vector<int>> f(n+1,vector<int>(m+1,0));
 9
10         for(int i=0;i<=n;i++)
11             f[i][0]=i;
12
13         for(int j=0;j<=m;j++)
14             f[0][j]=j;
15
16         for(int i=1;i<=n;i++)
17             for(int j=1;j<=m;j++)
18                 if(word1[i-1]==word2[j-1])
19                     f[i][j]=f[i-1][j-1];
20                 else{
21                     f[i][j]=min(f[i-1][j-1],min(f[i-1][j],f[i][j-1]))+1;
22
23                 }
24
25         return f[n][m];
26     }
27 };