A peak element is an element that is strictly greater than its neighbors.
Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.
You may imagine that nums[-1] = nums[n] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.
You must write an algorithm that runs in O(log n) time.
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: 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.
Constraints:
1 <= nums.length <= 1000-231 <= nums[i] <= 231 - 1nums[i] != nums[i + 1] for all valid i.
class Solution {
public int findPeakElement(int[] nums) {
if(nums.length==1){
return 0;
}else if ( nums[0]>nums[1]){
return 0;
}else if ( nums[nums.length-1]>nums[nums.length-2]){
return nums.length-1;
}else {
int start = 1 ;
int end = nums.length-2;
while(start<=end){
int mid = ( start + end)/2;
if(nums[mid]>nums[mid-1] && nums[mid]>nums[mid+1]){
return mid ;
}else if ( nums[mid]<nums[mid+1]){
start = mid+1 ;
}else{
end = mid-1 ;
}
}
return -1 ;
}
}
}
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int end=nums.size()-1;
int start=0;
int mid=start+ (end-start)/2;
while(start<end)
{
mid= start + (end-start)/2;
if(nums[mid]>nums[mid+1])end=mid;
else start=mid+1;
}
return end;
}
};
