L 280. Wiggle Sort ---M ~~~~可以允许直接两两交换

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

Example:

Input: nums = [3,5,2,1,6,4]
Output: One possible answer is [3,5,1,6,2,4]

A:

-------先来个 构造法 --------  排序，然后每2个位置，互换

class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        for(int i = 1;i+1<n;i+=2){
            //swich i, and i+1 position
            int tmp = nums[i];
            nums[i] = nums[i+1];
            nums[i+1] = tmp;
        }        
    }
};

然而上面的代码明显不是最优的。

考虑到我们允许  ==   因此可以直接反复交换

class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        int tmp=0;
        for(int i = 0; i+1 < nums.size(); i++){
            if(i&1){// position 1,3,5...
                if(nums[i] < nums[i+1]){
                    //swich i, and i+1 position
                    tmp = nums[i];
                    nums[i] = nums[i+1];
                    nums[i+1] = tmp;
                }
            }else{ // position 0 2,4,6,8
                if(nums[i] > nums[i+1]){
                    //swich i, and i+1 position
                    tmp = nums[i];
                    nums[i] = nums[i+1];
                    nums[i+1] = tmp;
                }
            }
        }        
    }
};