【LeetCode & 剑指offer刷题】查找与排序题9：Find Peak Element
【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
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.
Note:
Your solution should be in logarithmic complexity.
//问题：找极大值元素（返回任意一个极大值均可）
/*
方法一：找最大值一定是极大值
O(n)
*/
class Solution
{
public:
int findPeakElement(vector<int>& nums)
{
int n = nums.size();
if(n==0) return -1; //为空时，返回-1
int max_index = 0;
for(int i = 0; i < n; i++)//找序列中的最大值，一定是极大值
{
if(nums[i]>nums[max_index]) max_index = i;
}
return max_index;
}
};
/*掌握
方法二：借助二分查找思路
right指的数比后一个数大，left指的数比前一个数大，当两个“指针”相遇时，就可以满足极大值条件（两个指针按二分跨度走，故效率较高）
有点像lower_bound函数
O(logn)
*/
class Solution
{
public:
int findPeakElement(vector<int>& nums)
{
if(nums.empty() return -1; //为空时，返回-1
int left = 0,right = nums.size()-1;
while(left<right) //left = right时退出循环，结果是left,right,mid均指向极大值位置
{
int mid = (left+right)/2;
if(nums[mid] < nums[mid+1]) //如果中间值比右边值小，说明峰值在右边
left = mid+1;
else //如果中间值比右边值大，说明峰值在左边（包括a[mid]，故取right=mid）
right = mid;
}
return right;
}
};
