【LeetCode 75】Sort Colors

题目：

Given an array with n objects colored red, white or blue, sort them in-place 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.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

Follow up:

• A rather straight forward solution is a two-pass algorithm using counting sort.
  First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.
• Could you come up with a one-pass algorithm using only constant space?

1、使用系统的sort排序可通过。

2、计数排序

     计数排序适用于排序元素个数（种类）有限的情况下，如题目中给出数组中只有0,1,2三种数。

【思路】定义存储元素个数的一个一维数组--->再根据个数将元素放入原数组中

【代码】

class Solution {
    public void sortColors(int[] nums) {
        int[] count = new int[3];
        for(int i=0; i<nums.length; i++){
            //元素只有0,1,2
            count[nums[i]] ++;
        }
        int index = 0;
        for(int i=0; i<count[0]; i++){
            nums[index++] = 0;
        }
        for(int i=0; i<ccount[1]; i++){
            nums[index++] = 1;
        }
        for(int i=0; i<count[2]; i++){
            nums[index++] = 2;
        }
    }
}

3、三路快排

     适用于重复元素较多

class Solution {
    public void sortColors(int[] nums) {
        //初始值保证了数组是无效数组，没有数据，若zero=0，则nums[0]=0,但此时并不能确定nums数组中有元素0
        int zero = -1;            //nums[0...zero]=0
        int two = nums.length;   //nums[two...n-1]=2
        for(int i=0; i<two; ){
            if(nums[i]==1){
                i++;
            }
            else if(nums[i]==2){
                two--;
                swap(two, i, nums);
            }
            else{
                zero++;
                swap(zero, i, nums);
                i++;
            }
        }
    }
    public void swap(int i, int j, int[] nums){
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

https://www.imooc.com/article/16141