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.
思路:从头开始遍历,找到第一个A[i]大于A[i+1]的点,记peak=i,再从peak开始遍历,如果存在A[I]>A[peak],则说明这不是个山峰。
代码:
int peak;
int len=A.size();
for(int i=0;i<len-1;i++)
{
if(A[i]>A[i+1])
{
peak=i;
break;
}
}
for(int i=peak+1;i<len;i++)
{
if(A[peak]<A[i])
return -1;
}
return peak;