451. Sort Characters By Frequency

最新推荐文章于 2022-07-30 00:29:18 发布

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最新推荐文章于 2022-07-30 00:29:18 发布

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分类专栏： LeetCode Hash Table Heap 文章标签： LeetCode Hash Table Heap Sort Characters By F

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本文介绍了一种使用桶排序解决字符频率排序问题的方法。通过创建哈希表存储字符及其出现次数，并利用桶排序将字符按频率从高到低排序，最后构建出排序后的字符串。

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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.

用Bucket Sort解决问题。思路如下：

1、创建hashmap存相应的字符，

2、创建bucket在相应的框里存字符，

3、后向前遍历添加字符串。

代码如下：

public class Solution {
    public String frequencySort(String s) {
        HashMap<Character, StringBuilder> hm = new HashMap<Character, StringBuilder>();
        char[] chs = s.toCharArray();
        int len = chs.length;
        StringBuilder[] strs = new StringBuilder[len + 1];
        StringBuilder res = new StringBuilder();
        for (char ch: chs) {
            if (!hm.containsKey(ch)) {
                StringBuilder temp = new StringBuilder();
                hm.put(ch, temp);
            }
            hm.get(ch).append(ch);
        }
        for (char ch: hm.keySet()) {
            int lens = hm.get(ch).length();
            if (strs[lens] == null) {
                strs[lens] = new StringBuilder(); 
            }
            strs[lens].append(hm.get(ch));
        }
        for (int i = len; i >= 0; i--) {
            if (strs[i] != null) {
                res.append(strs[i]);
            }
        }
        return res.toString();
    }
}