Leetcode 451. Sort Characters By Frequency (用堆)

451. Sort Characters By Frequency
     Medium

Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string.

Return the sorted string. If there are multiple answers, return any of them.

Example 1:

Input: s = “tree”
Output: “eert”
Explanation: ‘e’ appears twice while ‘r’ and ‘t’ both appear once.
So ‘e’ must appear before both ‘r’ and ‘t’. Therefore “eetr” is also a valid answer.
Example 2:

Input: s = “cccaaa”
Output: “aaaccc”
Explanation: Both ‘c’ and ‘a’ appear three times, so both “cccaaa” and “aaaccc” are valid answers.
Note that “cacaca” is incorrect, as the same characters must be together.
Example 3:

Input: s = “Aabb”
Output: “bbAa”
Explanation: “bbaA” is also a valid answer, but “Aabb” is incorrect.
Note that ‘A’ and ‘a’ are treated as two different characters.

Constraints:

1 <= s.length <= 5 * 105
s consists of uppercase and lowercase English letters and digits.

解法1：

class Solution {
public:
    string frequencySort(string s) {
        priority_queue<pair<int, char>> maxHeap;
        unordered_map<char, int> mp;
        for (auto c : s) mp[c]++;
        for (auto m : mp) {
            maxHeap.push({m.second, m.first});
        }
        string res = "";
        while (!maxHeap.empty()) {
            auto top = maxHeap.top();
            maxHeap.pop();
            for (int i = 0; i < top.first; i++) res += top.second;   //注意不要写成res = res + top.second; 不然内存不够
        }
        return res;
    }
};
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解法2：

struct Node {
    char ch;
    int freq;
    Node(char c, int f) : ch(c), freq(f) {}
    bool operator < (const Node & n) const {
        return freq < n.freq;
    }
};

class Solution {
public:
    string frequencySort(string s) {
        int n = s.size();
        priority_queue<Node> maxHeap;
        unordered_map<char, int> char2Freq;
        for (auto c : s) {
            char2Freq[c]++;
        }
        for (auto cf : char2Freq) {
            maxHeap.push(Node(cf.first, cf.second));
        }
        string res = "";
        while (!maxHeap.empty()) {
            Node top = maxHeap.top();
            maxHeap.pop();
            for (int i = 0; i < top.freq; i++) res += top.ch; //注意不要写成res = res + top.second; 不然内存不够
        }
        return res;
    }
};
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