[leetcode] 280. Wiggle Sort 解题报告

题目链接: https://leetcode.com/problems/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].

思路: 遍历一遍数组, 如果是奇数位置并且其值比下一个大,则交换其值, 如果是偶数位置并且其值比下一个小, 则交换其值. 时间复杂度是O(N)

代码如下:

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

参考:https://leetcode.com/discuss/69508/c-o-n-concise-solution