72.Edit Distance

题目描述

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three 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’)

我的解法

解题思路

没想到面试里上来就来了一道hard题目,好在在提示下写出来了差不多。最主要就是状态转移方程代表的意义,dp[i][j]代表的是word1前i位转换为word2前j位需要最少的步骤。

实现代码

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class Solution {
public:
int minDistance(string word1, string word2) {
int n = word1.length();
int m = word2.length();
if (n == 0) return m;
if (m == 0) return n;
vector<vector<int>> dp(n+1, vector<int> (m+1,0));
for (int i = 0; i < n+1; i++)
dp[i][0] = i;
for (int j = 0; j < m+1; j++)
dp[0][j] = j;
for (int i = 1; i < n+1; i++){
for (int j = 1; j < m+1; j++){
if (word1[i-1] == word2[j-1]){
dp[i][j] = min(dp[i-1][j-1],min(dp[i-1][j]+1,dp[i][j-1]+1));
}
else{
dp[i][j] = min(dp[i-1][j-1]+1, min(dp[i-1][j]+1, dp[i][j-1]+1));
}
}
}
return dp[n][m];
}
};