题目
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 - 1such 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: 5Explanation: The largest mountain is [1,4,7,3,2] which has length 5.
Example 2:
Input: [2,2,2]Output: 0Explanation: There is no mountain.
Note:
0 <= A.length <= 100000 <= A[i] <= 10000
Follow up:
- Can you solve it using only one pass?
- Can you solve it in
O(1)space?
题意
数组中,如果一个区间的左半部分严格单调递增,右半部分严格单调递减,将这个区间称为一个“山”,要求在给定数组中求这样的区间的最大长度。
思路
用变量start、end分别表示这样的区间的两个端点,对于每一个start,找到能使区间成“山”的end值,计算长度,更新start并重复步骤。
代码实现
Java
class Solution {public int longestMountain(int[] A) {int ans = 0;int start = 0, end = 0;while (start < A.length) {if (end < A.length - 1 && A[end] < A[end + 1]) {while (end < A.length - 1 && A[end] < A[end + 1]) {end++;}if (end < A.length - 1 && A[end] > A[end + 1]) {while (end < A.length - 1 && A[end] > A[end + 1]) {end++;}ans = Math.max(ans, end - start + 1);}}start = end > start ? end : start + 1;end = start;}return ans;}}
