題目來源
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.

一看這題目，挺簡單的啊，最先想到的方法就是從前往后遍歷，遍歷到該元素比周圍元素大或者小的就返回。第一個和最后一個特殊處理一下。直接AC了。復雜度O(n)。

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int n = nums.size();
        if (n == 1 || nums[0] > nums[1])
            return 0;
        if (nums[n-1] > nums[n-2])
            return n-1;
        for (int i=1; i<n-1; i++) {
            if (nums[i] < nums[i-1] && nums[i] < nums[i+1])
                return i;
            if (nums[i] > nums[i-1] && nums[i] > nums[i+1])
                return i;
        }
        return -1;
    }
};

好吧，肯定有O(logN)的做法，看看去。
先來個O(n)的，但是簡潔的很啊==

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int n = nums.size();
        for (int i=1; i<n; i++) {
            if (nums[i] < nums[i-1])
                return i-1;
        }
        return n-1;
    }
};

還有O(logN)的，之前就沒想到可以用二分查找，這不是無序的嗎？怎么用二分呢？看看代碼就知道了。

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int n = nums.size();
        int l = 0, r = n - 1;
        int mid1 = 0, mid2 = 0;
        while (l < r) {
            mid1 = (l + r) / 2;
            mid2 = mid1 + 1;
            if (nums[mid1] < nums[mid2])
                l = mid2;
            else
                r = mid1;
        }
        return l;
    }
};

實際上也不難看懂，比如找到mid1 and mid2是遞增的，那么mid2及其后面一定有個極值點，因為最后一點是遞減的。同理mid1 and mid2是遞減的，那么mid1及其前面一定有個極值點，因為第一點是遞增的。