LeetCode笔记:451. Sort Characters By Frequency
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问题:
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.
大意:
给出一个字符串,基于其中字符出现的次数按照降序排列。
例1:
输入:
“tree”输出:
“eert”解释:
‘e’出现了两次,‘r’和‘t’都只出现了一次。
所以‘e’要在‘r’和‘t’的前面。因此“eetr”也是对的。例2:
输入:
“cccaaa”输出:
“cccaaa”解释:
‘c’和‘a’都出现了三次,所以“aaaccc”也是对的。
注意“cacaca”是错的,因为同样的字符必须在一起。例3:
输入:
“Aabb”输出:
“bbAa”解释:
“bbaA”也是对的,但“Aabb”不对。
注意‘A’和‘a’被视为两个不同的字符。
思路:
题目的要求是将字符按照出现次数排序,注意是区分大小写的,而且并没有说只有字母。
其实整个过程可以分为几步:
- 遍历收集不同字符出现的次数;
- 按照次数排序
- 组成结果数组,每个字符要出现对应的次数。
我用二维数组来记录字符和出现的次数,然后用冒泡法排序,最后组成数组。
这个做法能实现,但是时间复杂度太高了,在超长输入的测试用例中超时了。
代码(Java):
public class Solution { public String frequencySort(String s) { char[] arr = s.toCharArray(); String[][] record = new String[arr.length][2]; int num = 0; // 将字符串中的不同字符及其数量记录到二维数组中 for (int i = 0; i < arr.length; i++) { int j = 0; boolean hasFind = false; for (; j < num; j++) { if (arr[i] - record[j][0].charAt(0) == 0) { hasFind = true; int temp = Integer.parseInt(record[j][1]) + 1; record[j][1] = Integer.toString(temp); break; } } if (!hasFind) { record[j][0] = String.valueOf(arr[i]); record[j][1] = "1"; num ++; } } // 对二维数组按第二列排序 for (int i = 0; i < num; i++) { for (int j = 0; j < num-1; j++) { String[] temp = new String[2]; if (Integer.parseInt(record[j][1]) - Integer.parseInt(record[j+1][1]) > 0) { temp = record[j]; record[j] = record[j+1]; record[j+1] = temp; } } } // 结果 System.out.println(num); String result = ""; for (int i = num-1; i >= 0; i--) { System.out.println(record[i][1]); System.out.println(record[i][0]); for (int j = 0; j < Integer.parseInt(record[i][1]); j++) { result = result + record[i][0]; } } return result; }}
改进:
其实想了想ASCII码表总共就126个字符,可以像对待纯字母一样用一个数组来记录次数,同时用一个字符数组来记录对应位置的字符,在排序时跟着一起变就行了。
但是依然会超时,看来主要还是排序方式太慢了。
改进代码(Java):
public class Solution { public String frequencySort(String s) { char[] arr = s.toCharArray(); int[] map = new int[126]; char[] c = new char[126]; // 将字符串中的不同字符及其数量记录到数组中 for (char ch : s.toCharArray()) { map[ch]++; } // 对应字符 for (int i = 0; i < 126; i++) { c[i] = (char)i; } // 对次数排序 for (int i = 0; i < 126; i++) { for (int j = 0; j < 125; j++) { if (map[j] - map[j+1] > 0) { // 次数移动 int tempNum = map[j]; map[j] = map[j+1]; map[j+1] = tempNum; // 对应字符移动 char tempChar = c[j]; c[j] = c[j+1]; c[j+1] = tempChar; } } } // 结果 String result = ""; for (int i = 125; i >= 0; i--) { if (map[i] > 0) { for (int j = 0; j < map[i]; j++) { result = result + c[i]; } } } return result; }}
他山之石:
public class Solution { public String frequencySort(String s) { if (s == null) { return null; } Map<Character, Integer> map = new HashMap(); char[] charArray = s.toCharArray(); int max = 0; for (Character c : charArray) { if (!map.containsKey(c)) { map.put(c, 0); } map.put(c, map.get(c) + 1); max = Math.max(max, map.get(c)); } List<Character>[] array = buildArray(map, max); return buildString(array); } private List<Character>[] buildArray(Map<Character, Integer> map, int maxCount) { List<Character>[] array = new List[maxCount + 1]; for (Character c : map.keySet()) { int count = map.get(c); if (array[count] == null) { array[count] = new ArrayList(); } array[count].add(c); } return array; } private String buildString(List<Character>[] array) { StringBuilder sb = new StringBuilder(); for (int i = array.length - 1; i > 0; i--) { List<Character> list = array[i]; if (list != null) { for (Character c : list) { for (int j = 0; j < i; j++) { sb.append(c); } } } } return sb.toString(); }}
这个做法用map来记录字符出现的次数,比我的方法要快一些。然后创建数组,用序号来表示其出现的次数,省去了排序的过程,最后反过来得出结果就可以了。速度回快一些,但是属于空间换时间,在字符数量很大时会浪费很多空间。
合集:https://github.com/Cloudox/LeetCode-Record
版权所有:http://blog.csdn.net/cloudox_
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