[Lintcode]Find Peak Element 寻找峰值

最新推荐文章于 2021-09-15 01:00:00 发布

原创最新推荐文章于 2021-09-15 01:00:00 发布 · 319 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#lintcode

算法专栏收录该内容

191 篇文章

订阅专栏

介绍了一种使用二分法在特定整数数组中查找峰值的方法，并提供了一个Java实现示例，该方法能在O(log n)的时间复杂度内找到至少一个峰值元素的位置。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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)

二分法查找，添加布尔型返回值，让查找到true时可以马上返回停止继续查找。、

class Solution {
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        if(A.length < 3) return -1;
        
        int left = 0, right = A.length - 1;
        List<Integer> res = new ArrayList<Integer>();
        helper(A, left, right, res);
        
        return res.get(0);
    }
    
    boolean helper(int[] A, int left, int right, List<Integer> res) {
        if(left > right) return false;
        
        int mid = (left + right) / 2;
        if(mid - 1 >= 0 && mid + 1 <= A.length - 1 && A[mid - 1] < A[mid] &&
        A[mid + 1] < A[mid]) {res.add(mid); return true;};
        
        return helper(A, left, mid - 1, res) || helper(A, mid + 1, right, res);
        
        
    }
}