19.2.13 [LeetCode 72] Edit Distance
Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 operations permitted on a word:
- Insert a character
- Delete a character
- Replace a character
Example 1:
Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')
题意
求最小编辑距离
有3个操作:删除,插入和替换
题解
一开始想了个dp,不够快
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution { 2 public: 3 int minDistance(string word1, string word2) { 4 int l1 = word1.length(), l2 = word2.length(); 5 vector<vector<int>>dp(l1+1, vector<int>(l2+1, INT_MAX)); 6 for (int i = 0; i <= l1; i++)dp[i][0] = i; 7 for (int i = 0; i <= l2; i++)dp[0][i] = i; 8 for (int i = 1; i <= l1; i++) { 9 int minnum = INT_MAX; 10 for (int j = 1; j <= l2; j++) { 11 if (word1[i-1] == word2[j-1]) 12 dp[i][j] = dp[i - 1][j - 1]; 13 else { 14 int dp1 = dp[i - 1][j] + 1; 15 int dp2 = dp[i - 1][j - 1] + 1; 16 dp[i][j] = min(dp1, dp2); 17 if (minnum != INT_MAX) 18 dp[i][j] = min(minnum + j, dp[i][j]); 19 } 20 minnum = min(dp[i][j] - j, minnum); 21 } 22 } 23 return dp[l1][l2]; 24 } 25 };
其实是有个地方我没注意到: dp[i][j] 与 dp[i][j-1] 的关系实际上是映射着插入操作的
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution { 2 public: 3 int minDistance(string word1, string word2) { 4 int l1 = word1.length(), l2 = word2.length(); 5 vector<vector<int>>dp(l1+1, vector<int>(l2+1, INT_MAX)); 6 for (int i = 0; i <= l1; i++)dp[i][0] = i; 7 for (int i = 0; i <= l2; i++)dp[0][i] = i; 8 for (int i = 1; i <= l1; i++) { 9 for (int j = 1; j <= l2; j++) { 10 if (word1[i-1] == word2[j-1]) 11 dp[i][j] = dp[i - 1][j - 1]; 12 else { 13 dp[i][j] = min(dp[i - 1][j], min(dp[i - 1][j - 1], dp[i][j - 1])) + 1; 14 } 15 } 16 } 17 return dp[l1][l2]; 18 } 19 };
貌似换成数组会更快,我还是第一次知道可以不动态申请内存这样写数组……
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution { 2 public: 3 int minDistance(string word1, string word2) { 4 int l1 = word1.length(), l2 = word2.length(); 5 int dp[l1+1][l2+1]; 6 for (int i = 0; i <= l1; i++)dp[i][0] = i; 7 for (int i = 0; i <= l2; i++)dp[0][i] = i; 8 for (int i = 1; i <= l1; i++) { 9 for (int j = 1; j <= l2; j++) { 10 if (word1[i-1] == word2[j-1]) 11 dp[i][j] = dp[i - 1][j - 1]; 12 else { 13 dp[i][j] = min(dp[i - 1][j], min(dp[i - 1][j - 1], dp[i][j - 1])) + 1; 14 } 15 } 16 } 17 return dp[l1][l2]; 18 } 19 };
注定失败的战争,也要拼尽全力去打赢它;
就算输,也要输得足够漂亮。