[Swift]LeetCode845. 数组中的最长山脉 | Longest Mountain in Array

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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?

---

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：

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

---

156ms

class Solution {
    func longestMountain(_ A: [Int]) -> Int {
        var longest = 0
        var increasing = true
        var mountainLength = 0
        var hasReturn = false
        var i = 0 
        while  i < A.count - 1{
            if increasing {
                if A[i] < A[i+1] {
                    i += 1
                    mountainLength += 1
                }else if A[i] > A[i+1] {
                    increasing = false
                }else {
                    i += 1
                    mountainLength = 0
                }
            }else {
                if mountainLength == 0 {
                     if A[i] >= A[i+1] {
                        i += 1
                    }else {
                         increasing = true
                    }
                }else {
                    if A[i] > A[i+1] {
                        hasReturn = true
                        i += 1
                        mountainLength += 1
                    }else {
                        longest = max(longest, mountainLength + 1)
                        mountainLength  = 0
                        increasing = false
                    }
                }
            }
        }
        if hasReturn {
            return max(longest, mountainLength + 1)
        }else {
            return longest
        }
    }
}

---

Runtime: 164 ms

Memory Usage: 19.3 MB

class Solution {
    func longestMountain(_ A: [Int]) -> Int {
        var N:Int = A.count
        var res:Int = 0
        var up:[Int] = [Int](repeating:0,count:N)
        var down:[Int] = [Int](repeating:0,count:N)
        for i in stride(from:N - 2,through:0,by:-1)
        {
            if A[i] > A[i + 1]
            {
                down[i] = down[i + 1] + 1
            }
        }
        for i in 0..<N
        {
            if i > 0 && A[i] > A[i - 1]
            {
                up[i] = up[i - 1] + 1
            }
            if up[i] > 0 && down[i] > 0
            {
                res = max(res, up[i] + down[i] + 1)
            }
        }
        return res
    }
}

---

168ms

class Solution {
    func longestMountain(_ A: [Int]) -> Int {
        if (A.count < 3) {
            return 0
        }
        var l2r = Array(repeating: 0, count: A.count)
        var r2l = Array(repeating: 0, count: A.count)
        
        for i in 1..<A.count {
            if A[i] > A[i-1] {
                l2r[i] = l2r[i-1] + 1
            }
        }
        
        for i in stride(from: A.count - 2, through: 1, by: -1) {
            if A[i] > A[i+1] {
                r2l[i] = r2l[i+1] + 1
            }
        }
        
        var result = 0
        for i in 1..<A.count {
            if (l2r[i] > 0 && r2l[i] > 0) {
                result = max(result, l2r[i] + r2l[i] + 1)
            }
        }
        
        return result 
    }
}