845. Longest Mountain in Array

问题描述：

 

845. Longest Mountain in Array

DescriptionHintsSubmissionsDiscussSolution

 

Pick One

---

Let's call any (contiguous) subarray B (of A) a mountain if the following properties hold:

• B.length >= 3
• There exists some 0 < i < B.length - 1 such that B[0] < B[1] < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1]

(Note that B could be any subarray of A, including the entire array A.)

Given an array A of integers, return the length of the longest mountain. 

Return 0 if there is no mountain.

Example 1:

Input: [2,1,4,7,3,2,5]
Output: 5
Explanation: The largest mountain is [1,4,7,3,2] which has length 5.

Example 2:

Input: [2,2,2]
Output: 0
Explanation: There is no mountain.

Note:

1. 0 <= A.length <= 10000
2. 0 <= A[i] <= 10000

Follow up:

• Can you solve it using only one pass?
• Can you solve it in O(1) space?

解题思路：

我们可以用变量来标识是否出现峰值，以及出现峰值的最左边的下标

对于A[i] > A[i-1] ; A[i] < A[i-1]; A[i] == A[i-1]分别讨论。

1.A[i] > A[i-1]：

　　此时为上坡情景，注意首先出现时，要改变是否出现峰值的标志位和最左边界

2.A[i] < A[i-1]：

　　此时为下坡，可能会出现在一开始的场景，所以我们在对ret进行更新的时候，要注意峰值标志位以及最左边界是否是有效值。

3.A[i] == A[i-1]

　　一定要检查这种情况，若出现连续相同的值，则当前不管是上坡下坡还是与峰值同大，都将无效：设置l = -1， peak = -1;

代码：

class Solution {
public:
    int longestMountain(vector<int>& A) {
        int l = -1,ret = 0, peak = -1;
        for(int r = 1; r < A.size(); r++){
            if(A[r] > A[r-1]){
                if(peak != -1 || l == -1){
                    peak = -1;
                    l = r-1;
                }
            }else if(A[r] < A[r-1]){
                if(l != -1){
                    if(peak == -1) peak = 0;
                    ret = max(ret, r-l+1);
                }

            }else{
                l = -1;
                peak = -1;
            }
        }
        return ret;
    }
};