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?
此题难度系数很高，和Kth Largest Element in an Array有点关系，通过前一道题来求出中位数，然后遍历数组，大于中位数的数组值排在索引为奇数的位置（1，3，5...），小于中位数的数组值排在偶数的位置(...10,8,6,4,2,0)，将数组值等于中位数的值不变化（现在不理解这部分），代码如下：
public class Solution {
public void wiggleSort(int[] nums) {
int n = nums.length;
int mid = findKthlargest(nums,(n+1)/2);
int left = 0;
int right = nums.length-1;
int i = 0;
while(i<=right){
if(nums[newIndex(i,n)]>mid) swap(nums,newIndex(i++,n),newIndex(left++,n));
else if(nums[newIndex(i,n)]<mid) swap(nums,newIndex(i,n),newIndex(right--,n));
else i++;
}
}
public int findKthlargest(int[] nums,int k){
int n = nums.length;
int p = quickselect(nums,0,n-1,n-k+1);
return nums[p];
}
public int quickselect(int[] nums,int lo,int hi,int k){
int i = lo;
int j = hi;
int pivot = nums[hi];
while(i<j){
if(nums[i++]>pivot) swap(nums,--i,--j);
}
swap(nums,i,hi);
int order = i-lo+1;
if(order==k) return i;
else if(order>k) return quickselect(nums,lo,i-1,k);
else return quickselect(nums,i+1,hi,k-order);
}
public void swap(int[] nums,int i,int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
public int newIndex(int idx,int n){
return (1+2*idx)%(n|1);
}
}
详细的讲解在链接处：
https://discuss.leetcode.com/topic/41464/step-by-step-explanation-of-index-mapping-in-java