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LeetCode: 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.

Brute-force Solution

(find_peak_element.cpp) download
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class Solution {
public:
    int findPeakElement(const vector<int> &num) {
      int count = num.size();
      if (count == 1)
      {
          return 0;
      } else if (count > 1) {
          for (int i = 0; i < count; ++i){
              if (i >= 1 && i < count - 1 )
              {
                  if (num[i] > num[i-1] && num[i] > num[i+1])
                  {
                      return i;
                  }
              }
              if (i == 0)
              {
                  if (num[0] > num[1])
                  {
                      return 0;
                  }
              }
              if (i == count - 1)
              {
                  if (num[i] > num[i-1])
                  {
                      return count - 1;
                  }
                  }
         }
      }
    }
};

Time Complexity O(N) costs 40ms.

Divide & conquer Solution

(find_peak_element_dc.cpp) download
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class Solution {
public:
    int findPeakElement(const vector<int> &num) {
int n = num.size();
    int low = 0;
    int high = n - 1;
    int mid = 0;

    while ( low < high ) {
        mid = low + ( high - low ) / 2;
        if ( ( mid == 0 || num[mid] > num[mid-1] ) &&
                ( mid == n-1 ||  num[mid] > num[mid+1] )  ){
            return mid;
        }

        if (mid >0 && num[mid-1] > num[mid]){
            high = mid - 1;
        }else{
            low = mid + 1;
        }
    }

    return low;
    }
};

Time Complexity O(logN) costs 16ms.

Math proof from here