LeetCode852. Peak Index in a Mountain Array

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:

  1. 3 <= A.length <= 10000
  2. 0 <= A[i] <= 10^6
  3. 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;

猜你喜欢

转载自blog.csdn.net/qq_25406563/article/details/81321314