LeetCode #791 - Custom Sort String

最新推荐文章于 2022-11-15 17:41:07 发布

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最新推荐文章于 2022-11-15 17:41:07 发布

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本文深入探讨了如何根据一个已排序的字符串S，对另一个字符串T进行重新排列，以匹配S的排序规则。通过实例说明，如输入S为cba，T为abcd，输出cbad，解释了算法的实现过程。文章详细介绍了使用哈希表记录T中字符频率，再按S的顺序构建结果字符串的方法。

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题目描述：

S and T are strings composed of lowercase letters. In S, no letter occurs more than once.

S was sorted in some custom order previously. We want to permute the characters of T so that they match the order that S was sorted. More specifically, if x occurs before y in S, then x should occur before y in the returned string.

Return any permutation of T (as a string) that satisfies this property.

Example :

Input:

S = "cba"

T = "abcd"

Output: "cbad"

Explanation:

"a", "b", "c" appear in S, so the order of "a", "b", "c" should be "c", "b", and "a".

Since "d" does not appear in S, it can be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.

Note:

• S has length at most 26, and no character is repeated in S.

• T has length at most 200.

• S and T consist of lowercase letters only.

class Solution {
public:
    string customSortString(string S, string T) {
        unordered_map<char,int> hash;
        for(int i=0;i<T.size();i++)
        {
            if(hash.count(T[i])==0) hash[T[i]]=1;
            else hash[T[i]]++;
        }
        string result;
        for(int i=0;i<S.size();i++)
        {
            if(hash.count(S[i]))
            {
                for(int j=0;j<hash[S[i]];j++)
                    result+=S[i];
                hash.erase(S[i]);
            }
        }
        for(auto x: hash)
        {
            for(int j=0;j<x.second;j++)
                result+=x.first;
        }
        return result;
    }
};