[LeetCode] -- 451.Sort Characters By Frequency 解题报告_giving a string,sort it in decreasing-优快云博客

问题描述

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.

问题分析

该问题是要对字符串按照其中字符出现的次数对字符进行重新排序，并将其重新打印。所以需要对字符串进行以下处理：

统计字符串中每个字符出现的次数；
按照字符串中字符的出现的次数对字符进行排序；
按照顺序打印字符。

直观的思路是，先利用HashMap存储字符以及相应的次数，然后采用优先队列对其进行排序存储，最后输出组成新的字符串。此方法的时间复杂度为 $O(n)+O(m*\log{m})$ ，其中 $n$ 为字符串长度， $m$ 为不同的字符的个数。

关键问题

有了上述的思路，我们就可以进行代码实现了，此处对一些关键的代码进行分析：

利用HashMap统计字符出现的次数

Map<Character, Integer> map = new HashMap<>();
char[] ch = s.toCharArray();
for (char c : ch){
	map.put(c, map.getOrDefault(c, 0) + 1);
}

构建Tuple类型，同时存储字符和次数，并重写compareTo方法：

class Tuple implements Comparable<Tuple>{
	private final char character;
	private final int count;
	Tuple(char ch, int count){
		this.character = ch;
		this.count = count;
	}
	
	@Override
	public int compareTo(Tuple that) {
		return that.count - this.count;
	}
}

利用优先队列对有序对进行存储排序

PriorityQueue<Tuple> queue = new PriorityQueue<>();
// HashMap 遍历
for (HashMap.Entry<Character, Integer> entry : map.entrySet()){
   	// 插入元素，offer() 与 add() 方法的区别在于前者在失败时返回 false，而后者抛出异常
    queue.offer(new Tuple(entry.getKey(), entry.getValue()));
}
// 删除元素
// queue.poll();
// 头部查询
// queue.peek();