LeetCode 852. Peak Index in a Mountain Array
Solution1:我的答案
时间复杂度
class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
for (int i = 1; i < A.size() - 1; i++)
if (A[i] > A[i-1] && A[i] > A[i+1])
return i;
}
};
Solution2:
参考网址:https://leetcode.com/problems/peak-index-in-a-mountain-array/discuss/139891/c++-O(log-n)binary-serach
二分查找,时间复杂度
class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
int l = 0,u= A.size()-1;
int mid = (l+u)/2;
while( !(A[mid]>A[mid-1] && A[mid]>A[mid+1]) ){
if(A[mid]>A[mid-1]) l = mid+1;
else u=mid-1;
mid = (l+u)/2;
}
return mid;
}
};
应该是要用二分查找做,否则也太无聊了