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.
二分法:
class Solution{ public: int peakIndexInMountainArray(vector<int>& a){ int beg = 1,end = a.size(); int mid = (beg + end) / 2; while(beg <= end){ if(a[mid] < a[mid - 1]){ end = mid - 1; } else if(a[mid] < a[mid + 1]){ beg = mid + 1; } else{ break; } mid = (beg + end) / 2; } return mid; } };
最大值法
class Solution{ public: int peakIndexInMountainArray(vector<int>& a){ int max_elem = *max_element(a.begin(), a.end()); int pos; for(pos = 0; pos < a.size(); ++pos){ if(a[pos] == max_elem) break; } return pos; } };
原文地址:https://www.cnblogs.com/Mayfly-nymph/p/11273740.html
时间: 2024-10-10 02:04:43