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: 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 <= 100000 <= A[i] <= 10000
Follow up:
- Can you solve it using only one pass?
- Can you solve it in
O(1)space?
我们把数组 A 中符合下列属性的任意连续子数组 B 称为 “山脉”：
B.length >= 3- 存在
0 < i < B.length - 1使得B[0] < B[1] < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1]
（注意：B 可以是 A 的任意子数组，包括整个数组 A。）
给出一个整数数组 A，返回最长 “山脉” 的长度。
如果不含有 “山脉” 则返回 0。
示例 1：
输入：[2,1,4,7,3,2,5] 输出：5 解释：最长的 “山脉” 是 [1,4,7,3,2]，长度为 5。
示例 2：
输入：[2,2,2] 输出：0 解释：不含 “山脉”。
提示：
0 <= A.length <= 100000 <= A[i] <= 10000
156ms
1 class Solution {
2 func longestMountain(_ A: [Int]) -> Int {
3 var longest = 0
4 var increasing = true
5 var mountainLength = 0
6 var hasReturn = false
7 var i = 0
8 while i < A.count - 1{
9 if increasing {
10 if A[i] < A[i+1] {
11 i += 1
12 mountainLength += 1
13 }else if A[i] > A[i+1] {
14 increasing = false
15 }else {
16 i += 1
17 mountainLength = 0
18 }
19 }else {
20 if mountainLength == 0 {
21 if A[i] >= A[i+1] {
22 i += 1
23 }else {
24 increasing = true
25 }
26 }else {
27 if A[i] > A[i+1] {
28 hasReturn = true
29 i += 1
30 mountainLength += 1
31 }else {
32 longest = max(longest, mountainLength + 1)
33 mountainLength = 0
34 increasing = false
35 }
36 }
37 }
38 }
39 if hasReturn {
40 return max(longest, mountainLength + 1)
41 }else {
42 return longest
43 }
44 }
45 }
Runtime: 164 ms
Memory Usage: 19.3 MB
1 class Solution {
2 func longestMountain(_ A: [Int]) -> Int {
3 var N:Int = A.count
4 var res:Int = 0
5 var up:[Int] = [Int](repeating:0,count:N)
6 var down:[Int] = [Int](repeating:0,count:N)
7 for i in stride(from:N - 2,through:0,by:-1)
8 {
9 if A[i] > A[i + 1]
10 {
11 down[i] = down[i + 1] + 1
12 }
13 }
14 for i in 0..<N
15 {
16 if i > 0 && A[i] > A[i - 1]
17 {
18 up[i] = up[i - 1] + 1
19 }
20 if up[i] > 0 && down[i] > 0
21 {
22 res = max(res, up[i] + down[i] + 1)
23 }
24 }
25 return res
26 }
27 }
168ms
1 class Solution {
2 func longestMountain(_ A: [Int]) -> Int {
3 if (A.count < 3) {
4 return 0
5 }
6 var l2r = Array(repeating: 0, count: A.count)
7 var r2l = Array(repeating: 0, count: A.count)
8
9 for i in 1..<A.count {
10 if A[i] > A[i-1] {
11 l2r[i] = l2r[i-1] + 1
12 }
13 }
14
15 for i in stride(from: A.count - 2, through: 1, by: -1) {
16 if A[i] > A[i+1] {
17 r2l[i] = r2l[i+1] + 1
18 }
19 }
20
21 var result = 0
22 for i in 1..<A.count {
23 if (l2r[i] > 0 && r2l[i] > 0) {
24 result = max(result, l2r[i] + r2l[i] + 1)
25 }
26 }
27
28 return result
29 }
30 }
原文地址：https://www.cnblogs.com/strengthen/p/10590282.html
时间： 2024-08-29 02:23:19