72. Edit Distance

Posted 2020-08-15 warmland

ref: https://discuss.leetcode.com/topic/3136/my-o-mn-time-and-o-n-space-solution-using-dp-with-explanation/2

如果用dp[i][j]表示word1里[0,i]长的子串和word2[0,j]的最小编辑距离，那么状态转移方程应该这样构成：

dp[i][j] = dp[i-1][j-1], if word1[i] == word2[j]

dp[i][j] = max{dp[i-1][j], dp[i][j-1], dp[i-1][j-1]} + 1, otherwise

如果word1[i] == word2[j]的时候不需要解释，

如果不相等的时候可能有三种操作，使两个字符串对齐：

　　1. 替换字母：把word1[i] 替换成word2[j]，那么是dp[i-1][j-1]+1

　　2. 在Word1[i]后面强行插入一个word2[j]. 所以可能是dp[i][j-1]

　　3. 直接把word1[i]删了，dp[i-1][j]

 

所以用二维矩阵表示很容易：

 

 1 public int minDistance(String word1, String word2) {
 2     int len1 = word1.length();
 3     int len2 = word2.length();
 4     
 5     //distance[i][j] is the distance converse word1(1~ith) to word2(1~jth)
 6     int[][] distance = new int[len1 + 1][len2 + 1]; 
 7     for (int j = 0; j <= len2; j++)
 8         {distance[0][j] = j;} //delete all characters in word2
 9     for (int i = 0; i <= len1; i++)
10         {distance[i][0] = i;}
11 
12     for (int i = 1; i <= len1; i++) {
13         for (int j = 1; j <= len2; j++) {
14             if (word1.charAt(i - 1) == word2.charAt(j - 1)) { //ith & jth
15                 distance[i][j] = distance[i - 1][j - 1];
16             } else {
17                 distance[i][j] = Math.min(Math.min(distance[i][j - 1], distance[i - 1][j]), distance[i - 1][j - 1]) + 1;
18             }
19         }
20     }
21     return distance[len1][len2];        
22 }

但是，可以像uniquePath一样把二维压成一维

pre表示上同一行前一列的值，f[j-1]表示f[i-1][j-1]（因为pre没有更新进去），f[j]表示同一列上一行

 

 1     public int minDistance(String word1, String word2) {
 2         if(word1 == null || word2 == null) {
 3             return -1;
 4         }
 5         int len1 = word1.length();
 6         int len2 = word2.length();
 7         int[] distance = new int[len2 + 1];
 8         for(int i = 0; i <= len2; i++) {
 9             distance[i] = i;
10         }
11         for(int i = 1; i <= len1; i++) {
12             int pre = i;
13             for(int j = 1; j <= len2; j++) {
14                 int cur = 0;
15                 if(word1.charAt(i - 1) == word2.charAt(j - 1)) {
16                     cur = distance[j - 1];
17                 } else {
18                     cur = Math.min(pre, Math.min(distance[j], distance[j - 1])) + 1;
19                 }
20                 distance[j - 1] = pre;
21                 pre = cur;
22             }
23             distance[len2] = pre;
24         }
25         return distance[len2];
26     }