【Leetcode】Find Peak Element

题目链接：https://leetcode.com/problems/find-peak-element/
题目：

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.

Note:

Your solution should be in logarithmic complexity.

思路：

可能包含多个极值点(peak)，只要取任意一个就行。假设只有一个peak，用二分查找根据mid元素单增单减的关系可以判断peak在mid的左边还是右边。如果不止一个peak，用二分法划分也不会将查找区间[left,right]划到没有peak的区间。时间复杂度为O(logn)。
算法：

	public int findPeakElement(int[] nums) {
		int index = -1;
		int l = 0, h = nums.length - 1, m = 0;
		while (l <= h) {
			m = (l + h) / 2;
			int r = isPeakElement(nums, m);//判断i跟左右元素关系
			if (r == 0) {
				return m;
			} else if (r == -1) {//单减 peak在左边
				h = m - 1;
			} else if (r == 1) {//单增peak在右边
				l = m + 1;
			}
		}
		return index;
	}

	/**
	 * 判断i跟左右元素关系
	 * -1代表单减 0代表是peak 1代表单增
	 */
	public int isPeakElement(int[] nums, int i) {
		if (nums.length - 1 == i) { // 如果在最右边
			if (nums.length - 2 >= 0 && nums[nums.length - 2] < nums[nums.length - 1])//最后一个元素且比前一个元素大，是peak
				return 0;
			else if (nums.length - 2 < 0)//单元素肯定是peak
				return 0;
			else if (nums.length - 2 >= 0 && nums[nums.length - 2] < nums[nums.length - 1]) {
				return -1;
			}
		}
		if (0 == i) { // 如果在最左边
			if (i + 1 < nums.length && nums[i + 1] < nums[i])
				return 0;
			else if (i + 1 >= nums.length)
				return 0;
			else if (i + 1 < nums.length && nums[i + 1] > nums[i]) {
				return 1;
			}
		}
		// 否则左右都有数
		if (nums[i] > nums[i - 1]) {
			if (nums[i] > nums[i + 1])
				return 0;
			else
				return 1;
		} else {
			if (nums[i] > nums[i + 1])
				return -1;
			else
				return 1;
		}