算法分析與設計——LeetCode Problem.451 Sort Characters By Frequency

問題詳情

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Given a string, sort it in decreasing order based on the frequency of characters.

Example 1:

Input:
"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:
"cccaaa"

Output:
"cccaaa"

Explanation:
Both 'c' and 'a' appear three times, so "aaaccc" is also a valid answer.
Note that "cacaca" is incorrect, as the same characters must be together.

Example 3:

Input:
"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.

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問題分析及思路

題目大意是將一個字符串變爲一個按字母個數多少排序，同一個字母出現次數多排在前面的字符串。
由於個數不多，直接暴力破解，使用一個256位的數組進行遍歷即可，最後的算法時間打敗了百分之八十五的人！

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具體代碼

class Solution {
public:
    string frequencySort(string s) {
        int a[256];
        int howmany = 0;
        for(int i = 0; i < 256; i ++) a[i] = 0;
        for(int i = 0; s[i] != '\0'; i++) {
            if(a[s[i]] == 0) howmany++;
            a[s[i]]++;
        }
        string newS;
        while(howmany != 0) {
            int max = 0;
            int maxi = 0;
            int i;
            for(i = 0; i < 256; i++) {
                if(a[i] > max) {
                    max = a[i];
                    maxi = i;
                }
            }
            char c = maxi;

            for(int j = 0; j < max; j++) {
                newS += c;
            }
            a[maxi] = 0;
            howmany--;
        }
        return newS;
    }
};