[Swift]LeetCode324. 摆动排序 II | Wiggle Sort II

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

Example 1:

nums = [1, 5, 1, 1, 6, 4]
[1, 4, 1, 5, 1, 6]

Example 2:

nums = [1, 3, 2, 2, 3, 1]
[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?

---

给定一个无序的数组 nums，将它重新排列成 nums[0] < nums[1] > nums[2] < nums[3]... 的顺序。

示例 1:

nums = [1, 5, 1, 1, 6, 4]
[1, 4, 1, 5, 1, 6]

示例 2:

nums = [1, 3, 2, 2, 3, 1]
[2, 3, 1, 3, 1, 2]

说明:
你可以假设所有输入都会得到有效的结果。

进阶:
你能用 O(n) 时间复杂度和 / 或原地 O(1) 额外空间来实现吗？

---

380ms

class Solution {
    func wiggleSort(_ nums: inout [Int]) {
        
        if nums.count < 2 {
            return
        }
        
        var count = nums.count
        
        var begin = 0, end = count - 1, mid = (begin + end) >> 1
        
        func patition(_ begin : Int, _ end : Int) -> Int {
            let index = nums[begin]
            var low = begin , high = end
            while low < high {
                while low < high && nums[high] >= index {
                    high -= 1
                }
                if low < high {
                    nums[low] = nums[high]
                    low += 1
                }
                
                while low < high && nums[low] <= index {
                    low += 1
                }
                
                if low < high {
                    nums[high] = nums[low]
                    high -= 1
                }
                
            }
            
            nums[low] = index
            
            return low
        }
        
        var index = patition(begin, end)
        
        while index != mid {
            if index < mid {
                begin = index + 1
            }else {
                end = index - 1
            }
            
            index = patition(begin, end)
        }
        
        mid = nums[index]
        
        var n = count
        
        func A(_ i : Int) -> Int {
            return (1+2*i)%(count|1)
        }
        var i = 0, j = 0, k = count - 1
        while j <= k {
            if nums[A(j)] > mid {
                nums.swapAt(A(i), A(j))
                i+=1
                j+=1
            }else if nums[A(j)] < mid {
                nums.swapAt(A(j), A(k))
                k-=1
            }else {
                j+=1
            }
        }
    }
}