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  • LeetCode 852. Peak Index in a Mountain Array

    原题链接在这里:https://leetcode.com/problems/peak-index-in-a-mountain-array/

    题目:

    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:

    1. 3 <= A.length <= 10000
    2. 0 <= A[i] <= 10^6
    3. A is a mountain, as defined above.

    题解:

    When finding peak index using binary search, compare A[mid] and A[mid + 1].

    If A[mid] < A[mid + 1], peak must on right side of mid, l = mid + 1.

    Else, peak must be within mid left side including mid. r = mid.

    Time Complexity: O(logn). n = A.length.

    Space: O(1).

    AC Java:

     1 class Solution {
     2     public int peakIndexInMountainArray(int[] A) {
     3         if(A == null || A.length == 0){
     4             return -1;
     5         }
     6         
     7         int l = 0;
     8         int r = A.length - 1;
     9         while(l < r){
    10             int mid = l + (r - l) / 2;
    11             if(A[mid] < A[mid + 1]){
    12                 l = mid + 1;
    13             }else{
    14                 r = mid;
    15             }
    16         }
    17         
    18         return l;
    19     }
    20 }

    类似Find Peak Element.

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/12076035.html
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