【LeetCode 75】Sort Colors-蒲公英云

【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;
    }
}