LeetCode 324. Wiggle Sort II

题目

这道题目很有意思，有意思的是使用O(n)的时间效率和O(1)的空间效率解决。我会写一篇专业的博客来介绍一下

以下就是O(n)的时间效率和O(1)的空间效率。

class Solution {
public:
int n;
int findKth(vector<int>& nums, int s, int e, int k)
{
    int i = s + 1;
    int j = e;
    int l = s;
    while (i <= j)
    {
        while (i <= j && nums[i] < nums[l])
        {
            i++;
        }

        while (i <= j && nums[j] > nums[l])
        {
            j--;
        }

        if (i <= j)
        {
            swap(nums[i], nums[j]);
            i++;
            j--;
        }
    }

    swap(nums[l], nums[j]);

    if (j - l + 1 == k)
        return nums[j];
    else if (j - l + 1 < k)
    {
        return findKth(nums, j + 1, e, k - (j - l + 1));
    }
    else
        return findKth(nums, s, j - 1, k);
}

double findMid(vector<int>& nums)
{
    if (nums.size() & 1)
    {
        return findKth(nums, 0, nums.size() - 1, nums.size() / 2 + 1);
    }
    else
    {
        int x = findKth(nums, 0, nums.size() - 1, nums.size() / 2);
        int y = findKth(nums, 0, nums.size() - 1, nums.size() / 2 + 1);
        return 1.0 * (x + y) / 2;
    }
}

int getNext(int index)
{
    index -= 2;
    if (index < 0)
    {
        return n + index + (~n & 1) * (~index & 1);
    }

    return index;
}

int getPre(int index)
{
    index += 2;

    if (index >= n)
        return abs((n | 1) - index);

    return index;
}

int compare(int a, int b)
{
    if ((a ^ b) & 1)
    {
        return a & 1;
    }
    else
    {
        return a <= b;
    }
}

void wiggleSort(vector<int>& nums) {

    if (nums.size() <= 1)
        return;
    double mid = findMid(nums);
    
    int i, j, k;
    n = nums.size();
    i = j = (n & 1 ? n - 1 : n - 2);
    k = 1;
    while (compare(k, j))
    {
        if (nums[j] < mid)
        {
            swap(nums[i], nums[j]);
            i = getNext(i);
            j = getNext(j);
        }
        else if (nums[j] == mid)
        {
            j = getNext(j);
        }
        else
        {
            swap(nums[j], nums[k]);
            k = getPre(k);
        }
    }
}

};