0451. Sort Characters by Frequency

451. Sort Characters By Frequency #

题目 #

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.

题目大意 #

这道题是 Google 的面试题。

给定一个字符串,要求根据字符出现的频次从高到低重新排列这个字符串。

解题思路 #

思路比较简单,首先统计每个字符的频次,然后排序,最后按照频次从高到低进行输出即可。

代码 #


package leetcode

import (
	"sort"
)

func frequencySort(s string) string {
	if s == "" {
		return ""
	}
	sMap := map[byte]int{}
	cMap := map[int][]byte{}
	sb := []byte(s)
	for _, b := range sb {
		sMap[b]++
	}
	for key, value := range sMap {
		cMap[value] = append(cMap[value], key)
	}

	var keys []int
	for k := range cMap {
		keys = append(keys, k)
	}
	sort.Sort(sort.Reverse(sort.IntSlice(keys)))
	res := make([]byte, 0)
	for _, k := range keys {
		for i := 0; i < len(cMap[k]); i++ {
			for j := 0; j < k; j++ {
				res = append(res, cMap[k][i])
			}
		}
	}
	return string(res)
}


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