162. 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.
题意
这道题要求以log(n)的时间复杂度找到数组中的一个峰值
可以用二分查找处理
判断中点mid处是否满足nums[mid+1] > nums[mid]如果满足就向后找，如果不满足就向前找，可以至少找到一个峰值点
解法
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int left = 0;
int right = nums.size() - 1;
while (left < right)
{
int mid = left + (right - left) / 2;
if (nums[mid+1] > nums[mid])
left = mid+1;
else
right = mid;
}
return left;
}
};
