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?
思路:使用三个指针,red、white和blue;分别记录等于0和等于2的元素,当white移动到A[white]==0,则A[red]和A[white]交换,然后red和white向后移动,当A[white]==2,交换A[blue]和A[white],然后blue向前移动,除上两种情况外white向后移动。感觉有点像快速排序一样。
class Solution { public: void SwapColors(int &color1,int &color2) { int temp=color1; color1=color2; color2=temp; } void sortColors(int A[], int n) { if(A==NULL || n==0) return; int red=0; int white=0; int blue=n-1; while(white<=blue) { if(A[white]==0) { SwapColors(A[white],A[red]); white++; red++; } else if(A[white]==2) { SwapColors(A[white],A[blue]); blue--; } else { white++; } } } };