75. Sort Colors

最新推荐文章于 2019-07-05 17:50:17 发布

Krysta1

最新推荐文章于 2019-07-05 17:50:17 发布

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分类专栏：刷题文章标签： python leetcode

本文链接：https://blog.youkuaiyun.com/Krysta1/article/details/79634738

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本文介绍了一种一过性排序算法，该算法可以对红、白、蓝三种颜色的对象进行排序，使相同颜色的对象相邻，并按红、白、蓝的顺序排列。通过使用两个指针j和k，分别指向0和2应该出现的位置，可以在一次循环中完成排序。

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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?

我的思路：

就是一个排序算法。冒泡就可以了。

我的代码：

class Solution:
    def sortColors(self, nums):
        m = len(nums)
        for i in range(m - 1):
            for j in range(m - 1 - i):
                if nums[j] > nums[j + 1]:
                    nums[j], nums[j + 1] = nums[j + 1], nums[j]

Discuss：

void sortColors(int A[], int n) {
    int j = 0, k = n-1;
    for (int i=0; i <= k; i++) {
        if (A[i] == 0)
            swap(A[i], A[j++]);
        else if (A[i] == 2)
            swap(A[i--], A[k--]);
    }
}

学到：

一次循环就可以办到，一个从头计算一个个数，遇到0就和当前0应该在的位置交换。遇到2就从后面开始算。很巧妙。