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.

分析

这种找Peak Element的首先想到的是binary search, 因为有更优的时间复杂度，否则brute force O(n)的方法太直接了。

binary search的方法就是比较当前element与邻居，至于左邻居还是右邻居都可以，只要一致就行。不断缩小范围，最后锁定peak element.

这种类似题也有很多Follow up, 比如先增后减的数组怎么找指定元素，甚至先增后减再增的数组怎么找指定元素。

方法都是一样的，就是先得找到peak element的点，然后根据peak点将整个数组分成几部分，对于每个部分来说，是单调的，所以可以对每个部分分别用binary search来找元素。

复杂度

time: O(logn), space: O(1)

代码

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