花花酱 LeetCode 852. Peak Index in a Mountain Array

zxi — Sun, 17 Jun 2018 14:55:07 +0000

Problem

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.

Solution1: Liner Scan

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua
// Running time: 16 ms
class Solution {
public:
  int peakIndexInMountainArray(vector& A) {
    for (int i = 1; i < A.size(); ++i)
      if (A[i] < A[i - 1]) return i - 1;
  }
};

C++/STL

// Author: Huahua
// Running time: 16 ms
class Solution {
public:
  int peakIndexInMountainArray(vector& A) {
    return max_element(begin(A), end(A)) - begin(A);
  }
};

Solution 2: Binary Search

Find the smallest l such that A[l] > A[l + 1].

Time complexity: O(logn)

Space complexity: O(1)

C++

// Author: Huahua
// Running time: 18 ms
class Solution {
public:
  int peakIndexInMountainArray(vector& A) {
    int l = 0;
    int r = A.size();
    while (l < r) {
      int m = l + (r - l) / 2;
      if (A[m] > A[m + 1])
        r = m;
      else
        l = m + 1;
    }
    return l;
  }
};

Java

// Author: Huahua, 3 ms
class Solution {
  public int peakIndexInMountainArray(int[] A) {
    int l = 0;
    int r = A.length;
    while (l < r) {
      int m = l + (r - l) / 2;
      if (A[m] > A[m + 1])
        r = m;
      else
        l = m + 1;
    }
    return l;
  }
}

Python3

# Author: Huahua, 40 ms
class Solution:
  def peakIndexInMountainArray(self, A):
    l, r = 0, len(A)
    while l < r:
      m = l + (r - l) // 2
      if A[m] > A[m + 1]:
        r = m
      else:
        l = m + 1
    return l

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