LeetCode-72-Edit Distance
Posted 无名路人甲
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算法描述:
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‘)
解题思路:动态规划题。递推式为:
dp[i][j] = dp[i-1][j-1] if(word1[i-1]==word2[j-1])
dp[i][j] = min(min(dp[i][j-1], dp[i-1][j]), dp[])+1 if(word1[i-1]!=word2[j-1])
注意初始化问题。
int minDistance(string word1, string word2) { vector<vector<int>> dp(word1.size()+1,vector<int>(word2.size()+1,0)); dp[0][0] = 0; for(int i=1; i <= word1.size(); i++) dp[i][0]=i; for(int j=1; j <= word2.size(); j++) dp[0][j]=j; for(int i=1; i <= word1.size(); i++){ for(int j=1; j <= word2.size(); j++){ if(word1[i-1]==word2[j-1]){ dp[i][j] = dp[i-1][j-1]; }else{ dp[i][j] = min(min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1])+1; } } } return dp[word1.size()][word2.size()]; }
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