[Binary Search] Find Peak Element

1. Example
peak element in an array = > 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.
Imagine num[-1] = num[n] = -INF

s= [1,2,3,1]
Return index number 2 where 3 is a peak element


2. Implementation
Q1: peak element ? global or local?
A1: greater than its neighbors , local maximum
Q1: multiple peaks ?
A1: return any of them
Q1: how to find peak?
A1: [0] > [1] or not, [n-1] > [n-1] or not
i =1~ n-2
 [i] > [i-1] && [i] > [i+1]
[1,2,3,1]
i=0   1 > 2 X
i=1   2 >1 , 2 <3 X
i=2  3>2, 3 >1 O
i=3   1 > 3X

// Time:O(log(n))
public int findPeakElement(int[] nums)
{
      if (nums.length == 0|| nums == null)
         return -1;
 
     
      int l = 0; 
      int r = nums.length -1;
      while (l <= r)
      {
           if (l == r)
               return l;
           int m = (l+r) >>1;
           if (  nums[m] > nums[m+1] )   //3 >2
               r = m;
           else // 1 < 3
               l = m+1;
      }



      return l;


}

// Time: O(n) 
public int findPeakElement(int[] nums)
{


      // validate the input
      if (  nums == null || nums.length == 0)
         return -1;

      


      //for (int i = 0; i < nums.length -1; i++)
      //{
      //     if( i==0 && nums[i] > nums[i+1])
      //          return i;
      //     else if (i == num.length-1 && nums[i] > nums[i-1])
      //          return i;
      //     else if (i>0 && i < nums.length-1 && nums[i] > nums[i-1] && nums[i] > nums[i+1])
      //          return i;
      //}

       

      // NOTE: nums[0] as first comparison to save a iterating index and csn separate from index 1 to len-2
      int max = num[0];
      int index = 0;
      for (int i = 1; i < num.length -1; i++)
      {

            int cur = nums[i];
            int prev= nums[i-1];
            int next = nums[i+1];
            if ( cur > prev && cur > next && cur > max)
            {
                   max = cur;
                   index = i;
            }
      }




      if (max < nums[nums.length -1])
            return nums.length -1;

      

      return index;
      
}

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