Find Peak Element(LeetCode)

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

Given an input array nums, where nums[i] ≠ nums[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 nums[-1] = nums[n] = -∞.

Example 1:

Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:

Input: nums = [1,2,1,3,5,6,4]
Output: 1 or 5 
Explanation: Your function can return either index number 1 where the peak element is 2, 
             or index number 5 where the peak element is 6.

Note:

Your solution should be in logarithmic complexity.

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找到数组的一个局部最大值。要求时间复杂度O(lgn)。

首先最容易想到的是遍历，如果前一个数>小于后一个数，则返回。因为不满足是前一个数<=后一个数，如果这个数再>后后一个数，必然是局部极值。但是这个时间复杂度是O(n)。

class Solution {  
public:  
	int findPeakElement(const vector<int> &nums) {  
		for(int i=1;i<nums.size();i++){  
			if(nums[i-1]>nums[i])  
				return i-1;  
		}  
		return nums.size()-1;  
	}  
}; 

要求O(lgn)，自然想到二分法。如果中间数大于后边一个，那么右半部分肯定有一个局部最大值，别忘了nums[n]=负无穷；否则在左半部分同理。

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