题目描述
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.
click to show follow up.
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?
0表示左边的元素，1表示中间的，2表示右边的。
代码如下：
class Solution {
public:void sortColors(int A[], int n) {int left = 0;int right = n-1;for(int i=0;i<= right;i++){if(A[i]==2){swap(A[i],A[right]);right --;i --;}else if(A[i]==0){swap(A[i],A[left]);left ++;}}}
};
算法思想如下：
如果当前的元素是2，那么就和right指针进行交换，由于最后的那个位置已经确定了，那么我们迭代的次数就可以减少一次，因此for循环中写的是i<=right，
如果当前的原始是0，就和left指针进行交换，由于交换后的A[i]的元素如果不是本身的话就只能是1，就不用对i进行修改。
这里注意程序中第二个if用到的是else if，因为第一个if语句中我们可能对A[i]修改，两个if语句中只能执行一个。
