451. Sort Characters By Frequency

Problem Description:

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.

贴入代码及注释

char* frequencySort(char* s) {    long int a[128][2]; //二维数组a[i][0]为ascii码，a[i][1]为该符号出现的次数    int i,j,k;          //计数变量       int max[2]={0};     //max[0]存储出现次数最多的符号的次数，max[1]存储次数最多的符号    int count = 0;      //计算字符串长度不含/0    char* re;           //返回拼凑出的字符串指针    for(i=0;i<128;i++){ //将128ascii码复制，并把对应技术单位赋空    a[i][0]=i;    a[i][1]=0;}i=0;while(*(s+count)!='\0'){        //将每个字符串源码取出    a[*(s+count)][1]++;         //将取出的字符串进行次数统计，且对应字符串值即为二维数组下标    count++;                    //循环次数也即字符串长度去除/0}re = (char*)malloc((count+1) * sizeof(char));//开辟返回字符串空间k = 0;                                       //返回数组计数器初始化for(i=0;i<128;i++){                         //类似选择排序每次选出最最大次数的符号并存入返回数组for(j=0;j<128;j++){        if(a[j][1]>max[0]){        max[0] = a[j][1];        max[1] = j;    }    }    while( a[max[1]][1]--){                 //将对应字符的次数变换成相同字符填入返回数组    *(re+k) = a[max[1]][0];    k++;}    a[max[1]][1]=0;                        //次数赋空    max[1]=0;    max[0]=0;}*(re+k)='\0';return re;}

该题目注意点：示例中仅有小写字幕a-z，实际题目意思为所有字符号，所以最好将128ascii全部包含在内。

算法步骤

1、将二维数组第一变量a[i][0]表示ascii码 第二变量a[i][1]表示为对应的次数即计数用

2、循环取出字符串字符，并以字符为下标直接对计数单元进行计数(好处比循环进行字符比较再进行计数少了O（n）复杂度

a[*(s+count)][1]++;  

3、用选择排序的思想不断取出次数最大的字符并存入新数组，此时为两层循环第一层为总共循环次数，第二层为选出最大次数字符