LeetCode Sort Colors

LeetCode 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.

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?


这道题有意思。实质是要把由0,1,2组成的数组从小到大排序,本来用计数排序,分别统计一下0,1,2这3个数字各有多少个,然后重新构造一个排好序的数组即可,但是Follow up里提到不能使用two-pass算法,并且只能使用常数空间复杂度。

我稍加思考,想到一个可行的算法。因为数组中只有3个数字,那么排好序的数组开头肯定是0(如果有0的话),结尾肯定是2(如果有2的话)。所以我们只需一遍遍历,遇到0则放到开头,遇到2则放到结尾,遇到1的话,就先统计一下1的数目。最后把1重新赋值到所有0的后面,1的个数在前面已经统计过了,所以正好。

完整代码如下:

class Solution {
public:
	void sortColors(vector<int>& nums) {
		int p0 = 0, p2 = nums.size() - 1; // p0和p2分别代表下一个0和2应该填入的位置
		int n1 = 0; // 1的数量
		for (int i = 0; i <= p2; i++) {
			if (nums[i] == 0)nums[p0++] = nums[i];
			else if (nums[i] == 1)n1++;
			else {
				swap(nums[i], nums[p2]);
				p2--;
				i--;
			}
		}
		for (int i = 0; i < n1; i++)
			nums[p0++] = 1;
	}
};

本代码提交AC,用时3MS。

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