1092. Shortest Common Supersequence

Given two strings str1 and str2, return the shortest string that has both str1 and str2 as subsequences.  If multiple answers exist, you may return any of them.

(A string S is a subsequence of string T if deleting some number of characters from T (possibly 0, and the characters are chosen anywhere from T) results in the string S.)

Example 1:

Input: str1 = "abac", str2 = "cab"
Output: "cabac"
Explanation: 
str1 = "abac" is a subsequence of "cabac" because we can delete the first "c".
str2 = "cab" is a subsequence of "cabac" because we can delete the last "ac".
The answer provided is the shortest such string that satisfies these properties.

Note:

1. 1 <= str1.length, str2.length <= 1000
2. str1 and str2 consist of lowercase English letters.

class Solution {
    public String shortestCommonSupersequence(String s1, String s2) {
        int l1 = s1.length(), l2 = s2.length();
        int[][] dp = new int[l1 + 1][l2 + 1];
        for(int i = 1; i <= l1; i++) {
            for(int j = 1; j <= l2; j++) {
                if(s1.charAt(i - 1) == s2.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1] + 1;
                else {
                    dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
                }
            }
        }
        
        StringBuilder sb = new StringBuilder();
        while(l1 != 0 || l2 != 0) {
            if(l1 == 0) sb.insert(0, s2.charAt(--l2));
            else if(l2 == 0) sb.insert(0, s1.charAt(--l1));
            else if(s1.charAt(l1 - 1) == s2.charAt(l2 - 1)) {
                sb.insert(0, s1.charAt(--l1));
                --l2;
            }
            else if(dp[l1][l2] == dp[l1 - 1][l2]) {
                sb.insert(0, s1.charAt(--l1));
            }
            else if(dp[l1][l2] == dp[l1][l2 - 1]) {
                sb.insert(0, s2.charAt(--l2));
            }
        }
        return sb.toString();
    }
}

这题有意思啊，从largest common subsequence往前推

先把LCS的dp数组算出来，再通过判断来构造shortes common supersequence，

l1, l2是dp数组的index

具体的判断条件有：

1.l1 == 0, 说明l1已经添加完了，剩下的都是l2的，往里添加即可。注意因为dp数组是l1+1 * l2 + 1, 所以添加的时候要先 -- l2

2.l2 == 0, 说明l2已经添加完了，剩下的都是l1的，往里添加即可。注意因为dp数组是l1+1 * l2 + 1, 所以添加的时候要先 -- l1

下面用i j 代替了

3. s1.charAt(i - 1) == s2.charAt(j - 1), 把这个char insert进去，i--， j--

两个char不相等，那就看一下dp数组跳过了哪个char，把它insert进去

4. dp[i - 1][j] == dp[i][j], 说明是从上面过来的，跳过了s1.charAt(i - 1), 把它insert进去

5.dp[i][j - 1] == dp[i][j], 说明是从左边面过来的，跳过了s2.charAt(j - 1), 把它insert进去

最后返回

 https://www.youtube.com/watch?v=rfV2BJp8YA8

花哥救我狗命