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.
思路:
Solution (C++):
int peakIndexInMountainArray(vector<int>& A) {
int low = 0, high = A.size()-1;
while (low <= high) {
int mid = low + (high-low) / 2;
if (A[mid] > A[mid+1] && A[mid] > A[mid-1]) return mid;
else if (A[mid] < A[mid+1]) low = mid + 1;
else high = mid-1;
}
return -1;
}
性能:
Runtime: 16 ms Memory Usage: 7.1 MB
思路:
Solution (C++):
int peakIndexInMountainArray(vector<int>& A) {
for (int i = 0; i < A.size()-1; ++i) {
if (A[i] > A[i+1]) return i;
}
return -1;
}
性能:
Runtime: 12 ms Memory Usage: 7.4 MB
原文链接: https://www.cnblogs.com/dysjtu1995/p/12741443.html
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