LintCode Find Peak Element(查找峰值)

原题网址:http://www.lintcode.com/en/problem/find-peak-element/

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)

方法:二分法。答案可能不唯一,只需要得到一个即可。

class Solution {
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here
        int i = 1, j = A.length - 2;
        while (i <= j) {
            int m = (i + j) / 2;
            if (A[m - 1] < A[m] && A[m] > A[m + 1]) return m;
            if (A[m - 1] < A[m]) i = m + 1; else j = m - 1;
        }
        return i;
    }
}


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