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https://leetcode.com/problems/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
class Solution { public: int minDistance(string word1, string word2) { int m = word1.length(), n = word2.length(); vector<vector<int> > dp(m+1, vector<int> (n+1, 0)); for(int i=0;i<=m;++i) dp[i][0] = i; for(int j=0;j<=n;++j) dp[0][j] = j; for(int i=1;i<=m;++i) { for(int j=1;j<=n;++j) { if(word1[i-1] == word2[j-1]) { dp[i][j] = min(dp[i-1][j-1], dp[i-1][j]+1); } else { dp[i][j] = min(dp[i-1][j], min(dp[i][j-1], dp[i-1][j-1])) + 1; } } } return dp[m][n]; } };
https://leetcode.com/problems/distinct-subsequences/
Given a string S and a string T, count the number of distinct subsequences of T in S.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.
class Solution { public: int numDistinct(string s, string t) { int m = s.length(), n = t.length(); vector<vector<int> > dp(m+1, vector<int>(n+1, 0)); for(int i=0;i<=m;++i) dp[i][0] = 1; for(int i=1;i<=m;++i) { for(int j=1;j<=n;++j) { if(s[i-1] == t[j-1]) dp[i][j] = dp[i-1][j-1] + dp[i-1][j]; else dp[i][j] = dp[i-1][j]; } } return dp[m][n]; } };
leetcode@ [72/115] Edit Distance & Distinct Subsequences (Dynamic Programming)
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原文地址：http://www.cnblogs.com/fu11211129/p/5024022.html
