72. Edit Distance(js)
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:给定两个单词,从第一个单词转变为第二个单词最短需要几步,只能通过增删改其中的一个字符
代码如下:
/** * @param {string} word1 * @param {string} word2 * @return {number} */ var minDistance = function(word1, word2) { let m=word1.length,n=word2.length; // vector<vector<int>> dp(m+1,vector<int>(n+1,0)); let dp=[] for(let i=0;i<m+1;i++){ dp[i]=[]; for(let j=0;j<n+1;j++){ dp[i][j]=0; } } for(let i=1;i<=m;i++){ dp[i][0]=i; } for(let i=1;i<=n;i++){ dp[0][i]=i; } for(let i=1;i<=m;i++){ for(let j=1;j<=n;j++){ if(word1[i-1]==word2[j-1]) dp[i][j]=dp[i-1][j-1]; else{ dp[i][j]=Math.min(dp[i-1][j-1]+1,Math.min(dp[i-1][j]+1,dp[i][j-1]+1)); } } } return dp[m][n]; };