LeetCode BinarySearch 162. Find Peak Element

162. Find Peak Element

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.

solution 1: binary search

第一种binary search写法，当前一个数大于后一个数，说明处于下降段，hi = mid；当前一个数小于后一个数，说明处于上升段，lo = mid + 1。

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

solution 2 : binary search

分析mid与两侧数字的大小关系，总共有四种可能，分别考虑四种情况。

    public int findPeakElement(int[] nums) {
        if (nums.length == 1) return 0;
        if (nums.length == 2) return (nums[0] > nums[1]) ? 0 : 1;
        
        int lo = 0;
        int hi = nums.length - 1;
        
        while (lo  + 1 < hi) {
            int mid = lo + (hi - lo) / 2;
            
            if (nums[mid] < nums[mid - 1] && nums[mid] > nums[mid + 1]) {
                hi = mid; 
            } else if (nums[mid] > nums[mid - 1] && nums[mid] < nums[mid + 1]) {
                lo = mid;
            } else if (nums[mid] > nums[mid - 1] && nums[mid] > nums[mid + 1]) {
                return mid;
            } else if (nums[mid] < nums[mid - 1] && nums[mid] < nums[mid + 1]) {
                lo = mid;
            }
        }
        
        if (nums[lo] > nums[hi]) {
            return lo;
        } else return hi;
    }