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.
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?
代码:oj测试通过 Runtime: 47 ms
1 class Solution: 2 # @param A a list of integers 3 # @return nothing, sort in place 4 def sortColors(self, A): 5 # None case 6 if A is None: 7 return None 8 # two pointers: p0 for red and p2 for blue 9 p0 = 0 10 p2 = len(A)-1 11 i = 0 12 while i <= p2 : 13 if A[i] == 0 : 14 A[i],A[p0] = A[p0],A[i] 15 i += 1 16 p0 += 1 17 elif A[i] == 1 : 18 i += 1 19 else : 20 A[i],A[p2] = A[p2],A[i] 21 p2 -= 1
思路:
这道题的要求跟很多数组题的要求类似,如何遍历一遍数组就完成任务。
就是头尾各放两个指针。
因为有三种类型的元素,所以需要守住头尾两个指针就可以了。
很多数组题目都是利用交换元素值的技巧,遍历一次就可以了。