Leetcode 280: Wiggle Sort

最新推荐文章于 2020-02-28 01:37:51 发布

dolphin_wby

最新推荐文章于 2020-02-28 01:37:51 发布

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分类专栏： Leetcode 文章标签： Sort Array

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该博客介绍了如何解决LeetCode中的280题——摆动排序。主要内容包括理解题意，即对未排序数组进行原地重新排列，使得奇数位置的元素小于其后元素，偶数位置的元素大于其后元素，然后提供了一种根据定义进行排序的解决方案。算法的时间复杂度为O(n)，空间复杂度为O(1)。

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

For example, given nums = [3, 5, 2, 1, 6, 4], one possible answer is [1, 6, 2, 5, 3, 4].

Solution:

The rule is each element at odd index is smaller than the next element while element at even index is greater than the next. So we can sort the array according to the definition.

Time complexity: O(n)

Space complexity: O(1)

public class Solution {
    public void wiggleSort(int[] nums) {
        int length = nums.length;
        boolean odd = true;
        for (int i = 0; i < length - 1; i++) {
            if (odd) {
                if (nums[i] > nums[i + 1]) {
                    swap(nums, i, i + 1);
                }
                odd = !odd;
            } else {
                if (nums[i] < nums[i + 1]) {
                    swap(nums, i, i + 1);
                }
                odd = !odd;
            }
        }
    }
    
    private void swap (int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}