Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character
这个题目的解法是动态规划。其实操作a和b本质上是一样的,从word1中删除一个字符和从word2中插入一个字符实质上没有任何区别。
dp[i][j]表示word1中前i个字符与word2中前j个字符的Edit Distance。
状态转移方程:
if(word1[i] != word2[j])
dp[i][j] = min(dp[i - 1][j], //delete one character from word1
dp[i][j - 1], //delete one character from word2
dp[i - 1][j - 1] //replace one character from both)
else
dp[i][j] = dp[i - 1][j - 1]
Solution: Dynamic Programming
Time Complexity: O(mn) Space Complexity: O(mn)
public class Solution {
public int minDistance(String word1, String word2) {
int len1 = word1.length();
int len2 = word2.length();
int[][] dp = new int[len1 + 1][len2 + 1];
//Initialization
for(int i = 1; i <= len2; i++){
dp[0][i] = i;
}
for(int i = 0; i <= len1; i++){
dp[i][0] = i;
}
//DP
for(int i = 1; i <= len1; i++){
for(int j = 1; j <= len2; j++){
if(word1.charAt(i - 1) == word2.charAt(j - 1)){
dp[i][j] = dp[i - 1][j - 1];
}else{
dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
}
}
}
return dp[len1][len2];
}
}
面试中应该想到的Test Cases:
1. “”, a 2. a, c 3. aaa, a 4. abc, bc