leetcode [324]Wiggle Sort II

标签：order   ...   strong   sorted   with   follow   数组   hat   code   

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

Example 1:

Input: nums = [1, 5, 1, 1, 6, 4]
Output: One possible answer is [1, 4, 1, 5, 1, 6].

Example 2:

Input: nums = [1, 3, 2, 2, 3, 1]
Output: 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?

题目大意：

将数组变成为nums[0] < nums[1] > nums[2] < nums[3]....这种形式。

解法：

将数组排好顺序，从左到右偶数位填充比中位数大的数字，从右到左奇数位填充比中位数小的数字，剩下的全填充中卫数。

java:

class Solution {
    public void wiggleSort(int[] nums) {
        if (nums.length<=1) return;
        Arrays.sort(nums);
        int len=nums.length,mid=nums[len/2],low=1,high=(len%2==1?len-1:len-2);
        int[] ans=new int[len];
        Arrays.fill(ans,mid);
        for (int i=len-1;i>=0&&nums[i]>mid;i--,low+=2){ ans[low]=nums[i]; }
        for (int i=0;i<len&&nums[i]<mid;i++,high-=2){ ans[high]=nums[i]; }
        for(int i=0;i<len;i++) nums[i]=ans[i];
    }
}

leetcode [324]Wiggle Sort II

标签：order   ...   strong   sorted   with   follow   数组   hat   code   

原文地址：https://www.cnblogs.com/xiaobaituyun/p/10905941.html