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.
Your solution should be in logarithmic complexity.
思路：log time，所以用二分法。二分法无非是左值、中间值、右值的比较。
如果mid<mid-1,那么下一次判断left~mid-1;（此时mid相当于-∞)
如果mid<mid+1,那么下一次判断mid+1~right（此时mid相当于-∞)
class Solution { public: int findPeakElement(vector<int>& nums) { return binarySearch(nums,0,nums.size()-1); } int binarySearch(vector<int>& nums, int start, int end){ if(start == end) return start; if(start+1 == end){ if(nums[start]>nums[end]) return start; else return end; } int mid = start + (end-start)/2; if(nums[mid]<nums[mid-1]){ return binarySearch(nums,start,mid-1); } else if(nums[mid]<nums[mid+1]){ return binarySearch(nums,mid+1,end); } else{ return mid; } } };
162. Find Peak Element (Array; Divide-and-Conquer)
原文：http://www.cnblogs.com/qionglouyuyu/p/6591755.html