(Leetcode) Edit Distance

Given 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

Topic:

Given two strings, the minimum number of operands required to change s to T.

3 character operations are insert, delete, and replace, respectively.

Ideas:

Dynamic Planning Ideas:

Suppose Dp[i][j] represents the minimum number of operands required to convert a string ending with s[i] and a string ending with t[j], consider three operations, and then take the three minimum values:

1. Replacement:

Assuming S[i-1],t[j-1] is aligned, i.e. Dp[i-1][j-1] is known, then when S[i]==t[j], dp[i][j]=dp[i-1][j-1], otherwise, dp[i][j]=dp[i-1][j-1]+1.

2. Delete

Suppose S[i-1],t[j] is aligned, that is, Dp[i-1][j] is known, the extra s[i] needs to be deleted, the operand +1, then dp[i][j]=dp[i-1][j]+1.

3. Insert

Suppose S[i],t[j-1] is aligned, i.e. Dp[i][j-1] is known, it needs to be inserted in S s[i+1]=t[j] to match, operand +1, then dp[i][j]=dp[i][j-1]+1.

State transition equation:

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)

Initial value:

Dp[i][0]=i

Dp[0][j]=j

Complexity of:

Complexity of Time: O (m*n)

Space complexity: O (M*n)

Space optimization:

By the state transfer equation, dp[i][j] and dp[i-1][j-1],dp[i-1][j],dp[i][j-1], can be removed one dimension, leaving only Dp[j].

The dp[i-1][j] and dp[i][j-1] on the right of the equation can be changed directly to Dp[j] (old values) and Dp[j-1] (updated), and only Dp[i-1][j-1] is not recorded, and can be saved by a variable.

So space complexity: O (N)

Code:

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]; }};

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];}    ;

　　

(Leetcode) Edit Distance