[LeetCode] 1092. Shortest Common Supersequence

LeetCode刷题记录

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Description

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 anywherefrom T) results in the string S.)

Example 1:

Input: str1 = "abac", str2 = "cab"
Output: "cabac"
Explanation:
str1 = "abac" is a substring of "cabac" because we can delete the first "c".
str2 = "cab" is a substring 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.

 

思路

题意：构造一个字符串，使得其子序列同时包含有 str1 和 str2，要求这个字符串在满足要求情况下长度最短

题解：找出 str1 和 str2 的最长公共子序列，剩余不在最长公共子序列中的字符拼接到这个最长公共子序列中。

class Solution {
public:
    string shortestCommonSupersequence(string str1, string str2) {
        string res = "";
        int len1 = str1.size(), len2 = str2.size();
        int dp[len1 + 5][len2 + 5];
        memset(dp, 0, sizeof(dp));
        int i, j;
        for (i = 1; i <= len1; i++){
            for (j = 1; j <= len2; j++){
                if (str1[i - 1] == str2[j - 1])    dp[i][j] = dp[i - 1][j - 1] + 1;
                else    dp[i][j] = dp[i][j - 1] > dp[i - 1][j] ? dp[i][j - 1] : dp[i - 1][j];
            }
        }
        i = len1, j = len2;
        string common = "";
        while (dp[i][j]){
            if (dp[i][j] == dp[i - 1][j])   i--;
            else if (dp[i][j] == dp[i][j - 1])  j--;
            else    common += str1[i - 1], i--, j--;
        }

        reverse(common.begin(), common.end());

        int len3 = common.size();
        i = 0, j = 0;
        for (int k = 0; k < len3; k++){
            while (i < len1 && str1[i] != common[k]){
                res += str1[i++];
            }
            while (j < len2 && str2[j] != common[k]){
                res += str2[j++];
            }
            res += common[k];
            i++;
            j++;
        }
        while (i < len1){
            res += str1[i++];
        }
        while (j < len2){
            res += str2[j++];
        }
        return res;
    }
};