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?
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
我们可以先给数组排序，然后在做调整。调整的方法是找到数组的中间的数，相当于把有序数组从中间分成两部分，然后从前半段的末尾取一个，在从后半的末尾去一个，这样保证了第一个数小于第二个数，然后从前半段取倒数第二个，从后半段取倒数第二个，这保证了第二个数大于第三个数，且第三个数小于第四个数，以此类推直至都取完，参见代码如下：
class Solution {
public:
void wiggleSort(vector<int>& nums) {
int len = nums.size();
int i = (len + 1) / 2;
int j = len;
vector<int> tmp = nums;
sort(tmp.begin(), tmp.end());
for (int k = 0; k < len; k++){
nums[k] = (k & 1) == 1 ? tmp[--j] : tmp[--i];
}
}
};