[LintCode] Find Peak Element 求数组的峰值

 

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.

Notice

The array may contains multiple peeks, find any of them.

Have you met this question in a real interview?

 

 

Example

Given [1, 2, 1, 3, 4, 5, 7, 6]

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

Challenge

Time complexity O(logN)

 

LeetCode上的原题，请参见我之前的博客Find Peak Element。

 

解法一：

class Solution {
public:
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    int findPeak(vector<int> A) {
        int left = 0, right = A.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (A[mid] < A[mid + 1]) left = mid + 1;
            else right = mid;
        }
        return right;
    }
};

 

解法二：

class Solution {
public:
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    int findPeak(vector<int> A) {
        for (int i = 1; i < A.size(); ++i) {
            if (A[i] < A[i - 1]) return i - 1;
        }
        return A.size() - 1;
    }
};