324. Wiggle Sort II(难)
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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?
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];}}};
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