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 thatB[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:
0 <= A.length <= 10000
0 <= A[i] <= 10000
Follow up:
- Can you solve it using only one pass?
- Can you solve it in
O(1)
space?
class Solution { public int longestMountain(int[] A) { int n=A.length; if(n<3)return 0; int left=0;int right;int max=0; while(left<n-2){ //skip decending and equal array while(left<n-1 && A[left]>=A[left+1]){ left++; } right=left+1; //mountain up while(right<n-1 && A[right]<A[right+1]){ right++; } //mountain down while(right<n-1 && A[right]>A[right+1]){ right++; //update the max value max=Math.max(max,right-left+1); } left=right; } return max; } }
好耶