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 thatA[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.
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 }