## Find Peak Element

A peak element is an element that is greater than its neighbors.

Given an input array where

`num[i] ≠ num[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

`num[-1] = num[n] = -∞`

.
For example, in array

`[1, 2, 3, 1]`

, 3 is a peak element and your function should return the index number 2.
Note:

Your solution should be in logarithmic complexity.

### Analysis:

It is required to have log(N) complexity, for the single array, binary search is usually considered since it satisfied the log(N) requirement.Be careful with the boundary conditions, it is a pretty simple problem.

### Code(C++):

class Solution { public: void fp(int st, int ed, const vector<int> &num, int &idx){ if (st > ed || idx != -1){ return; } else { int mid = st + (ed - st)/2; if (mid == num.size()-1 && num[mid]>num[mid-1]){ idx = mid; return; } if (mid == 0 && num[mid]>num[mid+1]){ idx = mid; return; } if (num[mid]>num[mid-1] && num[mid]>num[mid+1]){ idx = mid; return; } fp(st,mid-1,num, idx); fp(mid+1,ed,num, idx); } } int findPeakElement(const vector<int> &num) { if (num.size() == 1){ return 0; } int idx = -1; fp(0, num.size()-1,num,idx); return idx; } };

### Code(Python):

class Solution: # @param num, a list of integer # @return an integer idx = -1 def findPeakElement(self, num): if len(num) == 1: return 0 self.fp(0, len(num)-1, num) return self.idx def fp(self, st, ed, num): if st > ed or self.idx != -1: return else: mid = st + (ed - st)/2; if mid == len(num)-1 and num[mid]>num[mid-1]: self.idx = mid return if mid == 0 and num[mid]>num[mid+1]: self.idx = mid return if num[mid]>num[mid-1] and num[mid]>num[mid+1]: self.idx = mid return self.fp(st,mid-1,num) self.fp(mid+1,ed,num)

Hi Yu, you are looking at both segments beside mid, and there is no guarantee that the first segment it goes down along will have the answer. So it might come back and then go down the second segment as well. So isn't the worst case time complexity O(n) then ?

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