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?
一刷
题解：
方法1， 最简单的做法, 排序，然后两边取
方法2, 先找到数组的中位数，用类似于215 find largest kth的方法，然后把小于中位数的放左边，大于的放右边，然后两边取，space complexity O(n)
方法3， 要达到空间复杂度为O(1)，将nums数组的下标x通过函数idx()从[0, 1, 2, ... , n - 1, n] 映射到 [1, 3, 5, ... , 0, 2, 4, ...]，得到新下标ix
public void wiggleSort(int[] nums) {
int median = findKthLargest(nums, (nums.length + 1) / 2);
int n = nums.length;
int left = 0, i = 0, right = n - 1;
while (i <= right) {
if (nums[newIndex(i,n)] > median) {
swap(nums, newIndex(left++,n), newIndex(i++,n));
}
else if (nums[newIndex(i,n)] < median) {
swap(nums, newIndex(right--,n), newIndex(i,n));
}
else {
i++;
}
}
}
private int newIndex(int index, int n) {
return (1 + 2*index) % (n | 1);
}
