README.md

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852. 山脉数组的峰顶索引

English Version

题目描述

给定一个长度为 n 的整数 山脉 数组 arr ，其中的值递增到一个 峰值元素 然后递减。

返回峰值元素的下标。

你必须设计并实现时间复杂度为 O(log(n)) 的解决方案。

 

示例 1：

输入：arr = [0,1,0]
输出：1

示例 2：

输入：arr = [0,2,1,0]
输出：1

示例 3：

输入：arr = [0,10,5,2]
输出：1

 

提示：

• 3 <= arr.length <= 105
• 0 <= arr[i] <= 106
• 题目数据 保证 arr 是一个山脉数组

解法

方法一

Python3

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        left, right = 1, len(arr) - 2
        while left < right:
            mid = (left + right) >> 1
            if arr[mid] > arr[mid + 1]:
                right = mid
            else:
                left = mid + 1
        return left

Java

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

C++

class Solution {
public:
    int peakIndexInMountainArray(vector<int>& arr) {
        int left = 1, right = arr.size() - 2;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr[mid] > arr[mid + 1])
                right = mid;
            else
                left = mid + 1;
        }
        return left;
    }
};

Go

func peakIndexInMountainArray(arr []int) int {
	left, right := 1, len(arr)-2
	for left < right {
		mid := (left + right) >> 1
		if arr[mid] > arr[mid+1] {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

TypeScript

function peakIndexInMountainArray(arr: number[]): number {
    let left = 1,
        right = arr.length - 2;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (arr[mid] > arr[mid + 1]) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

Rust

impl Solution {
    pub fn peak_index_in_mountain_array(arr: Vec<i32>) -> i32 {
        let mut left = 1;
        let mut right = arr.len() - 2;
        while left < right {
            let mid = left + (right - left) / 2;
            if arr[mid] > arr[mid + 1] {
                right = mid;
            } else {
                left = left + 1;
            }
        }
        left as i32
    }
}

JavaScript

/**
 * @param {number[]} arr
 * @return {number}
 */
var peakIndexInMountainArray = function (arr) {
    let left = 1;
    let right = arr.length - 2;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (arr[mid] < arr[mid + 1]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return left;
};