[LintCode] 75 Find Peak Element 解题报告

Description
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.

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)

思路
这道题的题意告诉我们，一定有peak。因为一开始是升序的，最后是降序的。
那么我们在任何一点进行判断，如果这个点的值比左边大，那边这个点右边一定有peak。反之，这个点左边一定有peak。
我们用二分法来做。

Code

class Solution {
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here
        
        if (A == null || A.length == 0) {
            return -1;
        }
        
        int start = 0;
        int end = A.length - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid - 1] < A[mid]) {
                start = mid;
            }
            else {
                end = mid;
            }
        }
        
        return A[start] > A[end] ? start : end;
    }
}