> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/binarysearch/162.-find-peak-element.md).

# 162. Find Peak Element

**分界点很多找其中任意一个的情况**。例如findPeak, 它的peak(波)可以有很多，peak(波)之间也没有性质能把它们区分出来，乍看上去不适用于二分。但题目要求并不是找特定的peak，而是找到任意一个就行了。因此给定的数列可以划分成俩坨，前面那坨满足严格上升（题有说前后俩元素不会相等），后面那坨满足严格下降，返回的也就是个划分点peak。只是我们无法用这个方法找到特定的peak。

由于要返回的是peak, 如果在判断上升后把start移到mid后可能会错过这最后的peak。因此判断不是上升把end移动到mid前是最合理的。还有当mid == start时，如\[2, 3]，因为现在不会让start移到mid后，需要强制退出。

```
/*
 * @lc app=leetcode id=162 lang=cpp
 *
 * [162] Find Peak Element
 */
class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        if (nums.size() == 1) return 0;
        int start = 1, end = nums.size() - 1;
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (start == mid) {
                start = nums[end] > nums[start] ? end : start;
                break;
            }
            if (!(nums[mid] > nums[mid - 1])) end = mid - 1;
            else start = mid;
        }
        return start == 1 && nums[0] > nums[start] ? 0 : start;
    }
};


```


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