Wiggle Sort 解答

Question

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 1

仔细观察题目，有个条件很重要就是nums[0] <= nums[1] >= nums[2] <= nums[3]....也就是说对于奇书的 i 我们要保证 nums[i - 1] <= nums[i] <= nums[i + 1] 所以第一个想法是1. 先排序 2. 交换每个nums[i] 和 nums[i + 1]

Time complexity O(n log n).

 1 public class Solution {
 2     public void wiggleSort(int[] nums) {
 3         if (nums == null || nums.length < 2)
 4             return;
 5         Arrays.sort(nums);
 6         for (int i = 1; i < nums.length - 1; i = i + 2) {
 7             int tmp = nums[i];
 8             nums[i] = nums[i + 1];
 9             nums[i + 1] = tmp;
10         }
11     }
12 }

Solution 2

分奇偶来思考。

如果i是奇数，那么nums[i]应该大于nums[i - 1]，否则交换

如果i是偶数，那么nums[i]应该小于nums[i - 1]，否则交换

 1 public class Solution {
 2     public void wiggleSort(int[] nums) {
 3         if (nums == null)
 4             return;
 5         for (int i = 1; i < nums.length; i++) {
 6             if (i % 2 == 1) {
 7                 if (nums[i] < nums[i - 1])
 8                     swap(nums, i, i - 1);
 9             } else {
10                 if (nums[i] > nums[i - 1])
11                     swap(nums, i, i - 1);
12             }
13         }
14     }
15     private void swap(int[] nums, int x, int y) {
16         int tmp = nums[x];
17         nums[x] = nums[y];
18         nums[y] = tmp;
19     }
20 }