Problem:
Given an array with n objects colored red, white or blue, sort them in-place 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.
Example:
Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]
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 a one-pass algorithm using only constant space?
思路:
利用快速排序的思想,以1为基准,将所有的0放到1的左边,将所有的2放到1的右边,即可完成排序。
Solution:
void sortColors(vector<int>& nums) {
int zero = 0, second = nums.size()-1;
for (int i = 0; i <= second; i++) {
while (nums[i] == 2 && i < second) swap(nums[i], nums[second--]);
while (nums[i] == 0 && i > zero) swap(nums[i], nums[zero++]);
}
}
性能:
Runtime: 0 ms Memory Usage: 8.6 MB