LeetCode Q324: Wiggle Sort II

Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....

Example:
(1) Given nums = [1, 5, 1, 1, 6, 4], one possible answer is [1, 4, 1, 5, 1, 6].
(2) Given nums = [1, 3, 2, 2, 3, 1], one possible answer is [2, 3, 1, 3, 1, 2].

Note:
You may assume all input has valid answer.

Follow Up:
Can you do it in O(n) time and/or in-place with O(1) extra space?

Solution:

Please refer to this post

	class Solution {
	public:
	void wiggleSort(vector<int>& nums) {
	int n = nums.size();
	int mid = n/2;
	nth_element(nums.begin(), nums.begin() + mid, nums.end());
	threeWayPartition(nums, nums[mid]);
	vector<int> res(n);
	int largeStart = n-1;
	int smallStart = (n%2) ? mid : (mid-1);
	for (int i = 0; i < n; i+=2)
	res[i] = nums[smallStart--];
	for (int i = 1; i < n; i+=2)
	res[i] = nums[largeStart--];
	nums = res;
	}
	// this ensures all values equal to the median is in the middle
	void threeWayPartition(vector<int> &nums, int val) {
	int i = 0, j = 0;
	int n = nums.size()-1;
	while (j <= n){
	if (nums[j] < val)
	swap(nums[i++], nums[j++]);
	else if (nums[j] > val)
	swap(nums[j], nums[n--]);
	else
	j++;
	}
	}
	};

view raw Q324.cpp hosted with ❤ by GitHub

常数空间的解法：

	void wiggleSort(vector<int>& nums) {
	if (nums.empty()) {
	return;
	}
	int n = nums.size();
	// Step 1: Find the median
	vector<int>::iterator nth = next(nums.begin(), n / 2);
	nth_element(nums.begin(), nth, nums.end());
	int median = *nth;
	// Step 2: Tri-partie partition within O(n)-time & O(1)-space.
	auto m = [n](int idx) { return (2 * idx + 1) % (n | 1); };
	int first = 0, mid = 0, last = n - 1;
	while (mid <= last) {
	//可以把Tri-partie好的数组想象成一个隐形的数组，
	//在Tri-partie好数组之后，就可以利用(2 * idx + 1) % (n | 1)来填数字了
	if (nums[m(mid)] > median) {
	swap(nums[m(first)], nums[m(mid)]);
	++first;
	++mid;
	}
	else if (nums[m(mid)] < median) {
	swap(nums[m(mid)], nums[m(last)]);
	--last;
	}
	else {
	++mid;
	}
	}
	}

view raw Q324Rnd3.cpp hosted with ❤ by GitHub