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) 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?
如果这个题是大于等于,那么这道题的条件宽松很多,只因为多了一个等号。由于等号的存在,当数组中有重复数字存在的情况时,也很容易满足题目的要求。
当i为奇数时,nums[i] >= nums[i - 1]
当i为偶数时,nums[i] <= nums[i - 1]
那么我们只要对每个数字,根据其奇偶性,跟其对应的条件比较,如果不符合就和前面的数交换位置即可。对于这道题:
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]; } }
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