颜色排序--Sort Colors
https://leetcode.com/problems/sort-colors/
Sort Colors
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?
题意：排序成红，白，蓝（0,1,2）
来自http://www.cnblogs.com/zuoyuan/p/3775832.html南郭子綦的解析
解题思路：这道题不允许使用排序库函数。那么最直观的解法是：遍历两遍数组，第一遍对0，1，2计数，第二遍对数组进行赋值，这样是可以ac的。但题目的要求是只使用常数空间，而且只能遍历一遍。那么思路就比较巧妙了。设置两个头尾指针，头指针p0指向的位置是0该放置的位置，尾指针p2指向的位置是2该放置的位置。i用来遍历整个数组，碰到0把它和p0指向的数交换，碰到2把它和p2指向的数交换，碰到1继续向后遍历。有点类似快速排序的分割数组这一步。
1 class Solution:
2 # @param {integer[]} nums
3 # @return {void} Do not return anything, modify nums in-place instead.
4 def sortColors(self, nums):
5 A=nums
6 if A==[]: return A
7 p0=0; p2=len(A)-1; i=0 #p0在头，p2在尾
8 while i<=p2:
9 if A[i]==2: #遇到2，将2和A[P2]的数互换，p2向前
10 A[i],A[p2]=A[p2],A[i] #此时不对i+=1,因为换过来的数依然需要判断
11 p2-=1
12 elif A[i]==0: #遇到0，将0和A[p0]的数互换，p0向后
13 A[i],A[p0]=A[p0],A[i]
14 p0+=1
15 i+=1 #此时i+=1,因为换的A[p0]要么是0要么是1，所以i继续向后
16 else:
17 i+=1 #遇到1,继续向后判断
18 #p0会前面的1进行修改换成0，不换的1排在中间成为最终结果
19 #得到的结果是按0,1,2顺序排序即要求的红，白，蓝
20
疑问：为什么最后不用return?
