LeetCode-Peak Index in a Mountain Array

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本文介绍了一种高效算法，用于在一维山形数组中查找峰顶元素的位置。通过运用二分查找法，该算法能在O(log n)的时间复杂度内完成任务，适用于长度在3到10000之间的数组。

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Description：
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.

题意：给定一个一维数组，要求找出位置i（0 < i < A.length - 1），另其满足 A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length - 1]；

解法：对于位置i左边和右边的元素来说，是已经排好序的了，因此我们可以利用二分查找法，定义left = 0, right = A.length - 1，mid = (left + right) /2 ；找到满足A[mid] > A[mid - 1] && A[mid] > A[mid + 1]的那个位置；

Java

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] && A[mid] > A[mid + 1]) return mid;
            else if (A[mid] > A[mid - 1]) left = mid + 1;
            else right = mid - 1;
        }
        return -1;
    }
}