【区间 dp】A022

一、Problem

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.)

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

二、Solution

方法一：dp

思路

先用 dp 求出 lms，因为题目要求结果串包含 s1 和 s2 的所有字符，而 lms 中可能包含不全，所以我们还要将没有出现在 lms 中的字符拼接到 ans 中…

class Solution { String LMS(char[] s1, char[] s2, int n, int m) { String f[][] = new String[n+1][m+1]; for (int i = 0; i <= n; i++) Arrays.fill(f[i], ""); for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { if (s1[i-1] == s2[j-1]) f[i][j] = f[i-1][j-1] + s1[i-1]; else f[i][j] = f[i-1][j].length() > f[i][j-1].length() ? f[i-1][j] : f[i][j-1]; } return f[n][m]; } public String shortestCommonSupersequence(String str1, String str2) { char[] s1 = str1.toCharArray(), s2 = str2.toCharArray(); int n = s1.length, m = s2.length; String lms = LMS(s1, s2, n, m); StringBuilder ans = new StringBuilder(); int i = 0, j = 0; for (char c : lms.toCharArray()) { while (i < n && s1[i] != c) ans.append(s1[i++]); while (j < m && s2[j] != c) ans.append(s2[j++]); ans.append(c); i++; j++; } ans.append(str1.substring(i)); ans.append(str2.substring(j)); return ans.toString(); } }

复杂度分析

时间复杂度：$O(n\times m\times k)$，k 是字符串拼接的时间空间复杂度：$O(n\times m)$，

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方法二：优化

dp 求 lms 那里频繁的字符串拼接是整个算法的瓶颈，所以可优化的地方在这

复杂度分析

时间复杂度：$O()$，空间复杂度：$O()$，