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.
A:------------------------check left part, if works, then stop, Otherwise, check right side ---------
class Solution { public: int findPeakElement(vector<int>& nums) { int n = nums.size(); if(n == 1 or nums[0]>nums[1]) return 0; if(nums[n-1] > nums[n-2]) return n-1; return helper(nums, 0,n-1); } private: int helper(vector<int>& nums, int start, int end)// start/end is valid, no need to check peak element { if(start+1 > end-1) return -1; int mid = start + (end-start)/2; if(nums[mid] > nums[mid-1] and nums[mid] > nums[mid+1]) return mid; int leftRes = helper(nums, start, mid); if(leftRes>=0) return leftRes; return helper(nums,mid, end); } };
Mistakes:
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