题目描述:

A peak element is an element that is greater than its neighbors.

Given an input array where num[i] ≠ num[i+1], find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

You may imagine that num[-1] = num[n] = -∞.

For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

**Note:**Your solution should be in logarithmic complexity.

二分搜索，每次比较找到的nums[mid]与邻居的大小来决定向哪边搜索,注意对于边界的处理,下标不能越界.

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class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        if(nums.size() == 1) return 0;
        if(nums.size() == 2) return nums[0] > nums[1] ? 0 : 1;
        int start = 0, end = nums.size(), mid, len = nums.size();
        while(start < end){
            mid = (start + end) / 2;
            if(mid == 0) return nums[0] > nums[1] ? 0 : 1;
            if(mid == nums.size() - 1) return nums[len - 2] > nums[len - 1] ? len - 2 : len - 1;
            if(nums[mid] > nums[mid - 1] && nums[mid] > nums[mid + 1]){
                return mid;
            }else if(nums[mid] > nums[mid - 1] && nums[mid + 1] > nums[mid]){
                start = mid + 1;
            }else{
                end = mid;
            }
        }
        return start;
    }
};