Find the maximum by binary search

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
= -∞

.

For example, in array 

[1, 2, 3, 1]

, 3 is a peak element and your function should return
the index number 2.

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Note:
Your solution should be in logarithmic complexity.
解答：二分查找找到局部最大值，而且二分查找能满足O(logn)
我的代码：
int findPeakElement(vector<int>& nums) {
int len = nums.size();
if(len==1)
return 0;
int beg = 0, end = len-1;
int mid = 0;
while(beg <= end){
mid = (beg+end)/2;
if(nums[mid-1]>nums[mid] && nums[mid]>nums[mid+1])
end = mid-1;
else if(nums[mid-1]<nums[mid] && nums[mid]<nums[mid+1])
beg = mid+1;
else if(nums[mid-1]<nums[mid] && nums[mid]>nums[mid+1])
return mid;
else if(nums[mid-1]>nums[mid] && nums[mid]<nums[mid+1]){
beg = mid+1;
}
}
return (beg+end)/2;
}

大牛的代码：
1.用递归
class Solution {
public:

int findPeakElement(const vector<int> &num) {
return Helper(num, 0, num.size()-1);
}
int Helper(const vector<int> &num, int low, int high)
{
if(low == high)
return low;
else
{
int mid1 = (low+high)/2;
int mid2 = mid1+1;
if(num[mid1] > num[mid2])
return Helper(num, low, mid1);
else
return Helper(num, mid2, high);
}
}
};

2.用迭代
class Solution {
public:
int findPeakElement(const vector<int> &num)
{
int low = 0;
int high = num.size()-1;

while(low < high)
{
int mid1 = (low+high)/2;
int mid2 = mid1+1;
if(num[mid1] < num[mid2])
low = mid2;
else
high = mid1;
}
return low;
}
};

3.简单但是O(n)的一遍遍历法
class Solution {
public:
int findPeakElement(const vector<int> &num) {
for(int i = 1; i < num.size(); i ++)
{
if(num[i] < num[i-1])
{// <
return i-1;
}
}
return num.size()-1;
}
};