852.Peak Index in a Mountain Array

最新推荐文章于 2025-09-10 07:15:00 发布

原创最新推荐文章于 2025-09-10 07:15:00 发布 · 223 阅读

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本文探讨了山形数组峰顶索引的快速查找算法，对比了线性搜索与折半查找两种方法。介绍了折半查找的递归与非递归实现，其时间复杂度从O(n)降低至O(log(n))，显著提升了效率。

Let's call an array A a mountain if the following properties hold:

A.length >= 3
There exists some 0 < i < A.length - 1 such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]

Given an array that is definitely a mountain, return any i such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1].

Example 1:

Input: [0,1,0]
Output: 1

Example 2:

Input: [0,2,1,0]
Output: 1

Note:

3 <= A.length <= 10000
0 <= A[i] <= 10^6
A is a mountain, as defined above.

class Solution {
    public int peakIndexInMountainArray(int[] A) {
        int i;
        for(i =0;i<A.length-1;i++){
            if(A[i]<A[i+1]) continue;
            else break;
        }
        return i;
        
    }
}

初看这道题，觉得这道题简单的离谱。但是，上述代码的效果并不好。

Runtime: 3 ms, faster than 38.95% of Java online submissions for Peak Index in a Mountain Array.

这种直接的方法的时间复杂度为O(n)。那么有没有更快的算法呢？显然是有的。

折半查找(递归）：

class Solution {
    public int peakIndexInMountainArray(int[] A) {
        return Middle(A,0,A.length -1);
    }
    
    public int Middle(int[] A, int left, int right){
        int mid = (left+right)/2;
        if(left != right){
             if(A[mid] > A[mid+1])
                return Middle(A,left,mid);
             else 
                return Middle(A,mid+1,right);
        } 
        else return mid;
    }
    
}

时间复杂度为O(log(n))。

折半查找（非递归）：

class Solution {
    public int peakIndexInMountainArray(int[] A) {
       int left = 0;
       int right = A.length - 1;
        while(left<right){
            int mid = (left + right)/2;
            if(A[mid]<A[mid+1])
                left = mid+1;
            else
                right = mid;
        }
        return left;
    }
    
}

这道题，看似简单，实际上，第一个想到折半查找的人并不多。