【Lintcode】075.Find Peak Element

题目:

There is an integer array which has the following features:

  • The numbers in adjacent positions are different.
  • A[0] < A[1] && A[A.length - 2] > A[A.length - 1].

We define a position P is a peek if:

A[P] > A[P-1] && A[P] > A[P+1]

Find a peak element in this array. Return the index of the peak.

Example

Given [1, 2, 1, 3, 4, 5, 7, 6]

Return index 1 (which is number 2) or 6 (which is number 7)

题解:

class Solution {
public:
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    int findPeak(vector<int> A) {
        if (A.size() < 3) {
            return -1;
        }
        
        int n = A.size();
        int start = 1;
        int end = n - 1;
        while (start < end - 1) {
            int mid = start + (end - start) / 2;
            if (A[mid] > A[mid - 1] && A[mid] > A[mid + 1]) {
                return mid;
            } else if (A[mid] > A[mid-1]) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if (A[start] > A[start + 1]) {
            return start;
        } else if (A[end] > A[end - 1]) {
            return end;
        }
        
        return -1;
    }
};
原文地址:https://www.cnblogs.com/Atanisi/p/6822290.html