LeetCode75 Sort Colors

问题描述

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.

click to show follow up.

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 an one-pass algorithm using only constant space?

简单分析

这道题目的解决思路很多，但是题目在最后说到能够使用 O(1) 的时间，所以就需要利用原来的数组解决这个问题。因为只有三个类别，可以考虑对数组元素进行交换来实现整个的分类。具体的解决方案如下：
1. 遍历数组
2. 如果为0，就从头往后放，如果为2，就从后往前放。这样放置完成以后中间剩下的就是1.

PS：在这里需要考虑对数组不能完全遍历。不然会影响到已分类好的2元素。具体的算法如下。

代码如下

public void sortColors(int[] nums) {    int rn =0,bn=0;//初始化红蓝的计数    //上限是nums.length-bn 因为后面的bn个是已经归类好的蓝色球，所以不需要进行操作    //如果继续进行操作会把归类好的打乱    for (int i = 0; i < nums.length-bn; i++) {        if(nums[i]==0){            //如果是红色            swap(nums,i,rn);            rn++;        }else if(nums[i]==2){            swap(nums,i,nums.length-1-bn);            bn++;            i--;        }    }}private void swap(int[] nums,int a,int b){    int temp = nums[a];    nums[a]=nums[b];    nums[b]=temp;}

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