75.Sort Colors

Tags : Medium Array Two Pointers Sort

Links: https://leetcode.com/problems/sort-colors/

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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?

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class Solution {
public:
    void sortColors(vector<int>& nums) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = nums.size();
        vector<int> store(3);
        for (int i = 0; i < n; ++i) ++store[nums[i]];

        int pos = 0;
        for (int i = 0; i < 3; ++i) {
            while (store[i]) {
                nums[pos++] = i;
                --store[i];
            }
        }
    }
};

Follow Up相当于提示了使用桶排序，一次遍历统计数量，第二次遍历写入。

---

class Solution {
public:
    void sortColors(vector<int>& nums) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = nums.size();
        int zero = 0, two = n - 1;
        for (int i = 0; i <= two; ++i) {
            if (nums[i] == 0)
                std::swap(nums[i], nums[zero++]);
            else if (nums[i] == 2)
                std::swap(nums[i--], nums[two--]);
        }
    }
};

一次遍历肯定想到的是双指针，一个指针zero指向首段，一个指针two指向尾端，从头开始遍历，遇到0就和zero交换，遇到2就和two交换。因为i始终不小于zero，所以无需额外操作，但是每次遇到2和two，可能存在原来two指向的就不符合顺序，所以此时需要让i后退一步来检查是否符合要求。