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.
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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:
给一个数组,代表被涂上红色,白色,蓝色的物体,排序使得同样颜色的挨在一起,不要使用内置的排序
red:0
white:1
blue:2
思考:
一个更直接的解法是用counting sort进行two pass的算法
首先,迭代这个数组,数0,1,2的数目,然后用这几个数目重写一下数组
思考如何用one pass的算法并且只使用常数空间
参考:
http://www.cnblogs.com/zuoyuan/p/3775832.html
思路:
首部设一个指针 p0,记录0的位置
尾部设一个指针 p2,记录2的位置
i 从头开始遍历,遇到2 则放在p2上,遇到0 则放在p0上
Python
class Solution(object):
def sortColors(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
if nums==[]:
return
p0=0
p2=len(nums)-1
i=0
while i<=p2:
if nums[i]==2:
nums[i],nums[p2]=nums[p2],nums[i]
p2-=1
elif nums[i]==0:
nums[i],nums[p0]=nums[p0],nums[i]
p0+=1
i+=1
else:
i+=1