leetcode Wiggle Sort II

leetcode 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?

题目地址： leetcode Wiggle Sort II

题意：

给定一个数组，要求进行如下排序： 奇数的位置要大于两边。 如 nums[1] > nums[0]  ，nums[1] > nums[2]

思路:

方法一

排序，然后两边分别取，复杂度O(nlogn)

注意排完序之后应该倒着来。比如[4,5,5,6]这个 数据。

Java 7ms

public class Solution {
    public void wiggleSort(int[] nums) {
        Arrays.sort(nums);
        int[] temp = new int[nums.length];
        int s = (nums.length + 1) >> 1, t = nums.length;
        for (int i = 0; i < nums.length; i++) {
            temp[i] = (i & 1) == 0 ?  nums[--s] : nums[--t] ;
        }

        for (int i = 0; i < nums.length; i++)
            nums[i] = temp[i];
    }
}

Python 104ms

class Solution(object):
    def wiggleSort(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        temp = sorted(nums)
        s, t = (len(nums) + 1) >> 1, len(nums)
        for i in xrange(len(nums)):
            if i & 1 == 0:
                s -= 1
                nums[i] = temp[s]
            else:
                t -= 1
                nums[i] = temp[t]

 

方法二

用快排的思想查找中位数，然后再合并两边。

最坏复杂度O(n^2)，平均复杂度O(n)

Java 6ms

public class Solution {
    public void wiggleSort(int[] nums) {
        int medium = findMedium(nums, 0, nums.length - 1, (nums.length + 1) >> 1);
        int s = 0, t = nums.length - 1 , mid_index = (nums.length + 1) >> 1;
        int[] temp = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] < medium)
                temp[s++] = nums[i];
            else if (nums[i] > medium)
                temp[t--] = nums[i];
        }

        while (s < mid_index) temp[s++] = medium;
        while (t >= mid_index) temp[t--] = medium;

        t = nums.length;
        for (int i = 0; i < nums.length; i++)
            nums[i] = (i & 1) == 0 ? temp[--s] : temp[--t];
    }

    private int findMedium(int[] nums, int L, int R, int k) {
        if (L >= R) return nums[R];
        int i = partition(nums, L, R);
        int cnt = i - L + 1;
        if (cnt == k) return nums[i];
        return cnt > k ? findMedium(nums, L, i - 1, k) : findMedium(nums, i + 1, R, k - cnt);
    }

    private int partition(int[] nums, int L, int R) {
        int val = nums[L];
        int i = L, j = R + 1;
        while (true) {
            while (++i < R && nums[i] < val) ;
            while (--j > L && nums[j] > val) ;
            if (i >= j) break;
            swap(nums, i, j);
        }
        swap(nums, L, j);
        return j;
    }

    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

 

本文是leetcode 324 Wiggle Sort II  的题解，更多题解可见

https://www.hrwhisper.me/leetcode-algorithm-solution/