[LeetCode] 451. Sort Characters By Frequency 解题报告

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.

这一题为medium难度，但是个人感觉比较简单，可以说的不多，很显然是遍历统计次数，然后输出即可，直接上思路。
方法：
遍历整个单词，对于每个字母统计出现的顺序。针对每个字母，我使用的是一个int[][]来存储。对于一个字符，int[0]存储的是它出现的次数，遇到一次就++，int[1]存储的是该字符的index，这是为了在后面对频率排序的时候，仍然知道对应频率的字符是哪一个。（其实这个地方就是一个key value对，但是为了在结果里面排名更靠前，使用array无疑是最快的）。
遍历完以后，针对int[0]降序排列，然后按照int[1]字符，int[0]的频率，输出即可，很简单。
下面是代码，复杂度为O(n)，实际运行时间大致在n到2n之间：

public class Solution {
public String frequencySort(String s) {
char[] cArr = s.toCharArray();
// int[][]={freq,index}
int[][] nArrAlphas = new int[128][2];
for (int i = 0; i < nArrAlphas.length; i++) {
nArrAlphas[i][1] = i;
}
for (char c : cArr) {
nArrAlphas[c][0]++;
}
Arrays.sort(nArrAlphas, new Comparator<int[]>() {

@Override
public int compare(int[] o1, int[] o2) {
return o2[0] - o1[0];
}
});
StringBuilder sb = new StringBuilder();
for (int i = 0; i < nArrAlphas.length; i++) {
if (nArrAlphas[i][0] > 0) {
for (int j = 0; j < nArrAlphas[i][0]; j++) {
sb.append((char) nArrAlphas[i][1]);
}
} else {
break;
}
}
return sb.toString();

}
}

实际上，在new int[][]的时候，可打印的字符应该不会多到128个，实际上应该少于这个值，我偷懒就直接写了128。