LintCode148: Sort Colors (三指针法)-优快云博客

本文深入探讨了三色排序问题的两种解决方案，一种是利用三指针法进行一次遍历，另一种是通过两次分区操作实现排序。文章详细解释了每种方法的实现细节和注意事项，为读者提供了清晰的算法思路。

Sort Colors
中文English
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.

Example
Given [1, 0, 1, 2], sort it in-place to [0, 1, 1, 2].

Challenge
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?

Notice
You are not suppose to use the library’s sort function for this problem.
You should do it in-place (sort numbers in the original array).

解法1：三指针法。
注意：当nums[p2]=2时，与nums[p3]交换，此时p3–，但p2不能++，因为交换过来的数可能是0，1，2，要在while循环里面继续处理。
代码如下：

class Solution {
public:
    /**
     * @param nums: A list of integer which is 0, 1 or 2 
     * @return: nothing
     */
    void sortColors(vector<int> &nums) {
        int n = nums.size();
        if (n == 0) return;
        int p1 = 0, p2 = 0, p3 = n - 1;
        while(p2 <= p3) {
            if (nums[p2] == 0) {
                swap(nums[p1], nums[p2]);
                p1++; p2++;
            } else if (nums[p2] == 1) {
                p2++;
            } else {
                swap(nums[p2], nums[p3]);
                p3--;
            }
        }
    }
};

解法2：比较土。就是来两次partition。第一次把0和1的分到左边，2的分到右边。第二次把0和1再分成左右两边。

class Solution {
public:
    /**
     * @param nums: A list of integer which is 0, 1 or 2
     * @return: nothing
     */
    void sortColors(vector<int> &nums) {
        int len = nums.size();
        int left = 0, right = len - 1;
        while (left <= right) {
            while (left <= right && nums[left] < 2) left++;
            while (left <= right && nums[right] == 2) right--;
            if (left <= right) {
                swap(nums[left], nums[right]);
                left++; right--;
            }
        }
        right = left - 1; left = 0;
        while (left <= right) {
            while (left <= right && nums[left] < 1) left++;
            while (left <= right && nums[right] == 1) right--;
            if (left <= right) {
                swap(nums[left], nums[right]);
                left++; right--;
            }
        }
        return;
    }
};