LeetCode-72.Edit Distance
Posted 月半榨菜
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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‘)
时间复杂度为O(mn)
public int minDistance(String word1, String word2) {//dp mytip int[][] dp = new int[word1.length()+1][word2.length()+1];//word1前i个字符和word2前j个字符最少的改变次数 for(int i=1;i<=word1.length();i++){//相当于word1和word2前同时加一个相同的字符,并初始化 dp[i][0] = i; } for(int i=1;i<=word2.length();i++){ dp[0][i]=i; } for(int i=0;i<word1.length();i++){ for(int j=0;j<word2.length();j++){ if(word1.charAt(i)==word2.charAt(j)){//如果i和j相同不需要操作 dp[i+1][j+1]=dp[i][j]; } else{//如果不同,判断增删改,哪个操作更少 int min = dp[i][j+1]>dp[i+1][j]?dp[i+1][j]:dp[i][j+1]; min = min>dp[i][j]?dp[i][j]:min; dp[i+1][j+1]=min+1; } } } return dp[word1.length()][word2.length()]; }
空间上可优化 只保存当前行和上一行
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