（LeetCode 72）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

题目：

给定两个字符串，求把S变成T所需的最小操作数。

3种字符操作分别为插入、删除、替换。

思路：

动态规划思想：

假设dp[i][j]表示以S[i]结尾的字符串和以T[j]结尾的字符串转换所需的最小操作数，考虑三种操作，然后取三者最小值：

1、替换：

假设S[i-1],T[j-1]已对齐，即dp[i-1][j-1]已知，则当S[i]==T[j]时，dp[i][j]=dp[i-1][j-1]，否则，dp[i][j]=dp[i-1][j-1]+1.

2、删除

假设S[i-1]，T[j]已对齐，即dp[i-1][j]已知，多出来的S[i]需删除，操作数+1，则dp[i][j]=dp[i-1][j]+1.

3、插入

假设S[i],T[j-1]已对齐，即dp[i][j-1]已知，需在S中插入S[i+1]=T[j]来匹配，操作数+1，则dp[i][j]=dp[i][j-1]+1.

状态转移方程：

dp[i][j]=min(dp[i-1][j-1]+(S[i]==T[j]?0,1),dp[i-1][j]+1,dp[i][j-1]+1)

初始值：

dp[i][0]=i

dp[0][j]=j

复杂度：

时间复杂度：O(m*n)

空间复杂度：O(m*n)

空间优化：

由状态转移方程可知，dp[i][j]与dp[i-1][j-1],dp[i-1][j],dp[i][j-1]有关，可以去掉一维，只留下dp[j]。

等式右边的dp[i-1][j]和dp[i][j-1]都可以直接改成dp[j]（旧的值）和dp[j-1]（已更新），只有dp[i-1][j-1]没有记录下来，通过某个变量保存起来之后就可以。

因此空间复杂度：O(n)

代码：

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class Solution { public:     int minDistance(string word1, string word2) {         int m=word1.length();         int n=word2.length();         vector<vector<int> > distance(m+1,vector<int>(n+1));                   for(int i=0;i<=m;i++){             for(int j=0;j<=n;j++){                 if(0==i){                     distance[i][j]=j;                 }                 else if(0==j){                     distance[i][j]=i;                 }                 else{                     distance[i][j]=min(distance[i-1][j-1]+((word1[i-1]==word2[j-1])?0:1),                                        min(distance[i-1][j]+1,distance[i][j-1]+1)                                        );                 }               }                }         return distance[m][n];     } };

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class Solution { public:     int minDistance(string word1, string word2) {         int m=word1.length();         int n=word2.length();         vector<int> distance(n+1);                   for(int i=0;i<=m;i++){             int last;             for(int j=0;j<=n;j++){                 if(0==i){                     distance[j]=j;                 }                 else if(0==j){                     last=distance[j];                     distance[j]=i;                 }                 else{                     int temp=distance[j];                     distance[j]=min(last+((word1[i-1]==word2[j-1])?0:1),                                        min(distance[j]+1,distance[j-1]+1)                                        );                     last=temp;                 }               }                }         return distance[n];     } };