描述
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.
法一：暴力搜索
class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
int maxNum = A[0];
int index = 0;
for(int i = 1; i < A.size(); i++)
{
if(maxNum<A[i])
{
index = i;
maxNum=A[i];
}
}
return index;
}
};
法二：寻找A[i]>A[i+1]的地方
class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
for(int i = 0;i+1<A.size();i++)
{
if(A[i]>A[i+1])
return i;
}
}
};
法三：二分查找
int peakIndexInMountainArray(vector<int> A) {
int l = 0, r = A.size() - 1, mid;
while (l < r) {
mid = (l + r) / 2;
if (A[mid] < A[mid + 1]) l = mid;
else if (A[mid - 1] > A[mid]) r = mid;
else return mid;
}
}
