第三周作业2(LeetCode75)

1. 题目描述(LeetCode75)

Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue.

Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note:
You are not suppose to use the library’s sort function for this problem.

2. 解决思路

先遍历一遍数组并分别统计出0，1，2的个数，然后再根据统计的数据按对等数量再用0，1，2依次覆盖数组。时间复杂度为O(n).

3. 完整代码

#include <stdio.h>#include <stdlib.h>int main(){   int* ColorArr = NULL;   int len = 0;   int countRed = 0;   int countWhite = 0;   int countBlue = 0;   printf("请输入颜色数组的长度：\n");   printf("（注意：只能输入0：红色，1：白色，2：蓝色，这三个代表颜色的整数！）\n");   scanf("%d" , &len);   while(len <= 0)   {       printf("长度必须大于0，请重新输入：\n");       scanf("%d" , &len);   }   ColorArr = (int*)malloc(sizeof(int)*len); //数组分配内存   printf("请输入用对应数字代表的不同颜色：\n");   for(int i = 0; i<len; i++)   {       scanf("%d" ,&ColorArr[i]);   }   //遍历数组，并记录下0,1,2的个数   for(int j = 0; j < len; j++)   {       if(ColorArr[j]==0){        countRed++;       }       else if(ColorArr[j]==1){        countWhite++;       }       else{        countBlue++;       }   }   //根据统计的数量分别按对等的数量依次对数组元素进行批量覆盖   for(int l = 0; l < countRed; l++)   {       ColorArr[l] = 0;   }   for(int m = countRed; m < countWhite + countRed; m++)   {       ColorArr[m] = 1;   }   for(int n = countWhite + countRed; n < len; n++)   {       ColorArr[n] = 2;   }   printf("排序后的颜色数组为:\n");    for (int k = 0; k < len; k++)    {        printf("%d ", ColorArr[k]);    }    scanf("%d", &len);  //加个输入让窗口停下来    free(ColorArr);  //释放内存    return 0;}