[LeetCode280] Wiggle Sort

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

For example, given nums = [3, 5, 2, 1, 6, 4], one possible answer is [1, 6, 2, 5, 3, 4].

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媽呀，這題完全想複雜了好麼，其實根本不用什麼sort，只需要correct the nums[i]’s position, which doesn’t satisfy: nums[i-1] < nums[i] > nums[i+1].

比如例子，我們一直check每個nums[i] 發現都滿足，直到我們知道了nums[3] = 1 doesnt satisfy nums[3] > nums[2]， 就把它們swap，爲什麼不會破壞前面的check？因爲我們已經check過nums[2] < nums[1], 現在我們有nums[3] < nums[2]. After swapping(nums[3], nums[2]) we still have nums[1] > nums[2]!

class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        for(int i = 0; i+1<nums.size(); ++i){
            if(((i%2) && nums[i] < nums[i+1]) || (!(i%2)) && nums[i] > nums[i+1]) swap(nums[i], nums[i+1]);
        }
    }
};

上面的code是跟後面的數比較，其實完全一樣的。