Sort Colors

题目：

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?

题意：
给定一个数组，其中数组的数只有0,1,2三种，分别表示的是红色，白色和蓝色。然后需要排序将原来数组中的元素按照红、白和蓝来排列，并且相同种颜色需要放在一起。然后返回重新调整后的数组。

题解：

LZ一开始采用最简单的方法，也就是做两次遍历，第一次是统计原来数组中各种颜色出现的次数；第二次遍历依次将上一次得到的结果重新分布在数组中。时间复杂度为O(n).

public class Solution
{
public void sortColors(int[] nums)
{
int length = nums.length;
if(length == 0 || nums == null)
return;
int red = 0;
int white = 0;
int blue = 0;
for(int i = 0; i < length; i++)
{
if(nums[i] == 0)
red++;
else if(nums[i] == 1)
white++;
else
blue++;
}
for(int i = 0; i < red; i++)
nums[i] = 0;
for(int j = red; j < red + white; j++)
nums[j] = 1;
for(int k = red + white; k < red + white + blue; k++)
nums[k] = 2;
}
}以上方法也是最简单的。但是需要遍历两次，而下面这种方法则是相当经典，只需遍历一次即可。它是采用平插法，也就是每一次看插入的是什么，如果是0，那么将表示0,1,2的三个数字的位置都要向后移动；如果是1，那么只需要将表示1和2的位置向后移动；如果是2，那么只需要将2的位置向后移动。
class Solution {
public:
void sortColors(int A[], int n) {
int i = -1;
int j = -1;
int k = -1;
for(int p = 0; p < n; p ++)
{
//根据第i个数字，挪动0~i-1串。
if(A[p] == 0)
{
A[++k] = 2; //2往后挪
A[++j] = 1; //1往后挪
A[++i] = 0; //0往后挪
}
else if(A[p] == 1)
{
A[++k] = 2;
A[++j] = 1;
}
else
A[++k] = 2;
}

}
}

非常巧妙，可以学习。