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Still a DP problem.
The transition function is not hard to get:
1. When s[i-1] == t[j-1], which means we dont have to do any thing to handle the current chars. Then dp[i][j] = dp[i-1][j-1].
2. Otherwise, find the minimum then plus 1, since each operation takes one step.
1 class Solution { 2 public: 3 int minDistance(string word1, string word2) { 4 int l1 = word1.size(), l2 = word2.size(); 5 if (l1 == 0) return l2; 6 if (l2 == 0) return l1; 7 vector<vector<int> > dp(l1+1, vector<int>(l2+1, 0)); 8 for (int i = 0; i <= l1; i++) dp[i][0] = i; 9 for (int i = 1; i <= l2; i++) dp[0][i] = i; 10 for (int i = 1; i <= l1; i++) { 11 for (int j = 1; j <= l2; j++) { 12 if (word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1]; 13 else dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1; 14 } 15 } 16 return dp[l1][l2]; 17 } 18 };
LeetCode – Refresh – Edit Distance
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原文地址：http://www.cnblogs.com/shuashuashua/p/4349416.html
