[LeetCode]98. SortColors颜色排序

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.

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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?

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解法１：显然直接排序是最简便的一种方法。不让使用库函数，懒得写排序算法直接上了。

class Solution {
public:
    void sortColors(vector<int>& nums) {
         sort(nums.begin(), nums.end());
    }
};

解法２：可以借助快速排序的划分思想，对输入做两次划分即可：第一次key=1，将所有<key的拿到数组前面；第二次key=2，将后半部分未排序数组所有<key的拿到后半部分数组的前面。

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int n = nums.size();
        if(n < 2) return;
        int pos = partition(nums, 0, n - 1, 1);
        partition(nums, pos, n - 1, 2);      
    }
private:
    int partition(vector<int>& nums, int low, int high, int key) {
        int n = nums.size();
        while(low < high) {
            while(low < n && nums[low] < key) ++low;
            while(high >= 0 && nums[high] >= key) --high;
            if(low < n && high >= 0 && low < high)
                swap(nums[low], nums[high]);
            ++low;
            --high;
        }
        return --low;
    }
};

解法３：一个更简便的方法是扫描输入，统计每个数出现的次数，然后再按次数写入这些数到输入里即可。

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int num[3] = {0, 0, 0};
        for(int i = 0; i < nums.size(); ++i) 
            ++num[nums[i]];
        int k = 0;
        for(int i = 0; i < 3; ++i)
            while(num[i]-- > 0) nums[k++] = i;
    }
};

解法４：初始化两个指针beg和end指向数组的头和尾，然后从头遍历数组，若遇到０，则交换当前值和beg所指值，且beg前移一步；若遇到１则继续；若遇到２则交换当前值和end所指值，且end后移一步。直到遍历到end指针处结束。

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int nums1 = 0, beg = 0, end = nums.size() - 1;
        for(int i = 0; i <= end; ++i) {
            if(nums[i] == 0) swap(nums[i], nums[beg++]);
            if(nums[i] == 2) swap(nums[i--], nums[end--]); //注意交换后当前位置还需要判断，需要--i
        }
    }
};