Question

852. 山脉数组的峰顶索引

English Version

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

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

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;
  }
}

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;
  }
};

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
}

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;
}

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
  }
}

/**
 * @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;
};