324. Wiggle Sort II

# Time:  O(nlogn)
# Space: O(n)

class Solution(object):
    def wiggleSort(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        nums.sort()
        mid = (len(nums) - 1) / 2
        nums[::2], nums[1::2] = nums[mid::-1], nums[:mid:-1]


# Time:  O(n) ~ O(n^2)
# Space: O(1)
# Tri Partition (aka Dutch National Flag Problem) with virtual index solution.
from random import randint


class Solution2(object):
    def wiggleSort(self, nums):
        """
        :type nums: List[int]
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        def nth_element(nums, n, compare=lambda a, b: a < b):
            def tri_partition(nums, left, right, target, compare):
                mid = left
                while mid <= right:
                    if nums[mid] == target:
                        mid += 1
                    elif compare(nums[mid], target):
                        nums[left], nums[mid] = nums[mid], nums[left]
                        left += 1
                        mid += 1
                    else:
                        nums[mid], nums[right] = nums[right], nums[mid]
                        right -= 1
                return left, right

            left, right = 0, len(nums)-1
            while left <= right:
                pivot_idx = randint(left, right)
                pivot_left, pivot_right = tri_partition(nums, left, right, nums[pivot_idx], compare)
                if pivot_left <= n <= pivot_right:
                    return
                elif pivot_left > n:
                    right = pivot_left-1
                else:  # pivot_right < n.
                    left = pivot_right+1

        def reversedTriPartitionWithVI(nums, val):
            def idx(i, N):
                return (1 + 2 * (i)) % N

            N = len(nums) / 2 * 2 + 1
            i, j, n = 0, 0, len(nums) - 1
            while j <= n:
                if nums[idx(j, N)] > val:
                    nums[idx(i, N)], nums[idx(j, N)] = nums[idx(j, N)], nums[idx(i, N)]
                    i += 1
                    j += 1
                elif nums[idx(j, N)] < val:
                    nums[idx(j, N)], nums[idx(n, N)] = nums[idx(n, N)], nums[idx(j, N)]
                    n -= 1
                else:
                    j += 1

        mid = (len(nums)-1)//2
        nth_element(nums, mid)
        reversedTriPartitionWithVI(nums, nums[mid])

Solution from kamyu104/LeetCode-Solutions · MIT