题目：
A peak element is an element that is greater than its neighbors.
Given an input array wherenum[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 thatnum[-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.
解答：
难度在于时间复杂度，需要是对数时间。又是搜索题，很自然会想到二分搜索。
我们来看中间位置发生的情况：
如果是1……2，3，2……9，那么很自然可以返回3所在的位置作为一个峰值点；如果是1……2，3，4……9，那么可以肯定峰值点出现在后半部分：3，4……9中；如果是1……4，3，2……9，那么可以肯定峰值点出现在前半部分：1……4，3中；如果仅剩两个数，峰值点在更大的数的位置；如果仅剩一个数，就返回其位置
class Solution {public:int findPeakElement(vector<int>& nums) {int range = nums.size() – 1;if((range/2 – 1 >= 0) && (range/2 + 1 <= range)){if((nums[range/2] > nums[range/2 – 1]) && (nums[range/2] > nums[range/2 + 1])){return (range/2);}else if(nums[range/2] < nums[range/2 – 1]){vector<int> tmp(nums.begin(), nums.begin() + range/2);return findPeakElement(tmp);}else if(nums[range/2] < nums[range/2 + 1]){vector<int> tmp(nums.begin() + range/2, nums.end());return (range/2 + findPeakElement(tmp)); // 注意基础位置的偏移量}}else{if(range == 0) return 0;if(range == 1){if(nums[0] > nums[1]) return 0;elsereturn 1;}}}};
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