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.
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?
解法一很直观，只要扫描一遍nums，把里面有多少个0,1,2记录下来就行了。
OJ要求one-pass，参考了喜刷刷的解法。假设当前的情况如下：
0......0 1......1 x1 x2 .... xm 2.....2
| | |
left cur right
1. 如果 x1 == 1， cur ++
2. 如果 x1 == 0, x1和nums[left]交换, left ++, cur ++
3. 如果 x1 == 2, x1和nums[right]交换， right --。
注意第三种case 不能 cur++, xm有可能是2或者0。但是和case2中x1和nums[left]交换之后不用担心这个问题。因为一开始cur = left, 如果cur领先了left，那么一定是case1发生了，这时nums[left] 一定等于1。
这种思路只适用于3种颜色的情况，换句话说如果颜色多于三种，那么必须用two-pass才能解。
class Solution(object):
def sortColors(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
n = len(nums)
left = 0
right = n - 1
i = 0 while i <= right:
if nums[i] == 0:
nums[i], nums[left] = nums[left], nums[i]
left += 1
i += 1
elif nums[i] == 2:
nums[i], nums[right] = nums[right], nums[i]
right -= 1
else:
i += 1