Find Peak Element

With $nums[-1] = nums[n] = -\infty$, find an index let $nums[i - 1] < nums[i]$ and $nums[i + 1] < nums[i]$. Note that for any $i$, $nums[i] \neq nums[i + 1]$. A linear solution is simple, but we can solve this problem in $O(\log N)$. Make use of the conditions that $nums[i] \neq nums[i + 1]$, we take a iteration way that is very similar to binary search. For an index $i$, we search peak element in the left side of it if $nums[i - 1] > nums[i]$, conversely, we search peak element in the right side.

To prove this algorithm is simple. If $nums[i - 1] > nums[i]$, as each adjoined element is different, there must be a $-1 \le j < i - 1$ that let $nums[j] < nums[i]$, so there must be a peek element in the range of $[0, i - 1)$. There are a similar conclusion if $num[i + 1] > nums[i]$ and in the right side.

public class FindPeakElement {

    public int findPeakElement(int[] nums) {
        int l = 0;
        int r = nums.length - 1;
        int mid;

        while(true) {
            mid = (l + r) / 2;
            if (mid - 1 >= 0 && nums[mid - 1] > nums[mid]) {
                r = mid - 1;
            } else if (mid + 1 < nums.length && nums[mid + 1] > nums[mid]) {
                l = mid + 1;
            } else {
                return mid;
            }
        }
    }
}