Find Peak element

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.

这道题也是用二分法，如果num[mid] < num[mid+1] 说明曲线在上升趋势， peak应该在后面 所以把start指针挪到mid+1上

如果num[mid] >num[mid+1], 说明是在下降的趋势， peak在左侧 把end指针指向mid

//O(log(n))
public class Solution {
    public int findPeakElement(int[] nums) {
        int start = 0;
  int end = nums.length - 1;
  int mid;
  while (start + 1 < end) {
   mid = start + (end - start) / 2;
   if (nums[mid] < nums[mid - 1]) {
    end = mid;
   } else if (nums[mid] < nums[mid + 1]) {
       start = mid;
   } 
   else {
    return mid;
   }
  }
  if (nums[start] > nums[end]) {
   return start;
  }
  return end;

    }
}
//O(n)
public class Solution {
    public int findPeakElement(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        if (nums.length == 1) {
            return 0;
        }
        for (int i = 0; i < nums.length; i++) {
            if (i == 0 && nums[i] > nums[i + 1]) {
                return i;
            }
            if (i == nums.length - 1 && nums[i - 1] < nums[i]) {
                return i;
            }
            if ( i > 0 && i < nums.length - 1 && nums[i] > nums[i - 1] && nums[i] > nums[i + 1]) {
                return i;
            }
        }
        return - 1;
    }
}