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852. 山脉数组的峰顶索引

题目描述

符合下列属性的数组 arr 称为 山脉数组 ：

  • arr.length >= 3
  • 存在 i（0 < i < arr.length - 1）使得：
    • arr[0] < arr[1] < ... arr[i-1] < arr[i]
    • arr[i] > arr[i+1] > ... > arr[arr.length - 1]

给你由整数组成的山脉数组 arr ，返回满足 arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1] 的下标 i 。

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

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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/**
 * @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;
};

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