75. Find Peak Element 【medium】

75. Find Peak Element 【medium】

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

• It's guaranteed the array has at least one peak.
• The array may contain multiple peeks, find any of them.
• The array has at least 3 numbers in it.

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)

 

解法一：

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

分类讨论。

 

解法二：

class Solution {
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here
        int start = 1, end = A.length-2; // 1.答案在之间，2.不会出界 
        while(start + 1 <  end) {
            int mid = (start + end) / 2;
            if(A[mid] < A[mid - 1]) {
                end = mid;
            } else if(A[mid] < A[mid + 1]) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if(A[start] < A[end]) {
            return end;
        } else { 
            return start;
        }
    }
}

参考http://www.jiuzhang.com/solution/find-peak-element/的解法，此法更简单。