[Swift]LeetCode42. 接雨水 | Trapping Rain Water

時間 2019-11-11

標籤 swift leetcode42 leetcode 雨水 trapping rain water 欄目 Swift 简体版

原文原文鏈接

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.git

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!github

Example:算法

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

給定 n 個非負整數表示每一個寬度爲 1 的柱子的高度圖，計算按此排列的柱子，下雨以後能接多少雨水。編程

上面是由數組 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度圖，在這種狀況下，能夠接 6 個單位的雨水（藍色部分表示雨水）。感謝 Marcos 貢獻此圖。數組

示例:微信

輸入: [0,1,0,2,1,0,1,3,2,1,2,1]
輸出: 6

【雙指針】12ms

咱們可能會想到在一次迭代中作某種方式，而不是單獨計算左右部分。
從圖中的動態編程方法來看，只要注意 $right_max [i] > left_max [i]$

算法app

Initialize $\text{left}left pointer to 0 and \text{right}right pointer to size-1$
While $\text{left}< \text{right}left<right, do:$
- If $\text{height[left]}height[left] is smaller than \text{height[right]}height[right]$
- If $height[left] >= left_max$
  - Else add $\text{ans}ans$
  - Add 1 to $\text{left}left.$
- Else

 1 class Solution {
 2     func trap(_ height: [Int]) -> Int {
 3         var leftMax = 0, rightMax = 0
 4         var leftPointer = 0, rightPointer = height.count - 1
 5         var trappedWater = 0
 6         
 7         while leftPointer < rightPointer {
 8             if height[leftPointer] < height[rightPointer] {
 9                 if height[leftPointer] > leftMax {
10                     leftMax = height[leftPointer]
11                 } else {
12                     trappedWater += leftMax - height[leftPointer]
13                 }
14                 leftPointer += 1
15             } else {
16                  if height[rightPointer] > rightMax {
17                     rightMax = height[rightPointer]
18                 } else {
19                     trappedWater += rightMax - height[rightPointer]
20                 }
21                 rightPointer -= 1
22             } 
23         }
24         return trappedWater
25     }
26 }

【動態編程】 16msthis

 1 class Solution {
 2   func trap(_ height: [Int]) -> Int {
 3     var leftMax = [Int]()
 4     var rightMax = [Int]()
 5     var maxL = 0
 6     var maxR = 0
 7     for i in 0..<height.count {
 8       if maxL < height[i] {
 9         maxL = height[i]
10       }
11       leftMax.append(maxL)
12       if maxR < height[height.count - 1 - i] {
13         maxR = height[height.count - 1 - i]
14       }
15       rightMax.append(maxR)
16     }
17     rightMax.reverse()
18     var result = 0
19     for i in 0..<height.count {
20       let wall = max(min(leftMax[i], rightMax[i]), height[i])
21       result += wall - height[i]
22     }
23     return result
24   }
25 }

【暴力破解】28msspa

 1 class Solution {
 2     func trap(_ height: [Int]) -> Int {
 3         if height.count <= 0 {
 4             return 0
 5         }
 6         var maxL = height[0]
 7         var rights : Array = Array<Int>(repeating: 0, count: height.count)
 8         var result = 0
 9         var maxR = 0
10         
11         for i in height.enumerated().reversed() {
12             if height[i.offset] > maxR {
13                 maxR = i.element
14                 rights[i.offset] = maxR
15             }else {
16                 rights[i.offset] = maxR
17             }
18         }
19         
20         for i in 0..<height.count {
21             if height[i] > maxL {
22                 maxL = height[i]
23             }
24             result += max(min(maxL, rights[i]) - height[i],0)
25         }
26         
27         
28         return result
29     }
30 }

【暴力破解】44ms

 1 class Solution {
 2     func trap(_ height: [Int]) -> Int {
 3         guard height.count > 2 else { return 0 }
 4         var res: Int = 0
 5         var leftMax = Array(repeating: 0, count: height.count)
 6         var rightMax = Array(repeating: 0, count: height.count)
 7     
 8         leftMax[0] = height[0]
 9         for i in 1..<height.count {
10             leftMax[i] = max(leftMax[i - 1], height[i])
11         }
12         
13         rightMax[height.count - 1] = height[height.count - 1]
14         for i in (0..<height.count - 1).reversed() {
15             rightMax[i] = max(rightMax[i + 1], height[i])
16         }
17     
18         for i in 1..<height.count - 1 {
19             res += min(leftMax[i], rightMax[i]) - height[i]
20         }
21         
22         return res
23     }
24 }

【使用堆棧】64ms

 1 class Solution {
 2     func trap(_ height: [Int]) -> Int {
 3         var stack = Stack<Int>()
 4         
 5         var ans = 0
 6         
 7         var current = 0
 8         
 9         while current < height.count {
10             while !stack.isEmpty && height[current] > height[stack.peek()] {
11                 let top = stack.pop()
12 
13                 if stack.isEmpty {
14                     break
15                 }
16                 
17                 let distance = current - stack.peek() - 1
18                 
19                 let height = min(height[current], height[stack.peek()]) - height[top]
20                 
21                 ans += distance * height
22             }
23             stack.push(current)
24             current += 1
25         }
26         
27         return ans
28     }
29 }
30 
31 class Stack<T> {
32     private var arr = [T]()
33     
34     var isEmpty: Bool {
35         return arr.isEmpty
36     }
37     
38     func peek() -> T {
39         return arr.last!
40     }
41     
42     func push(_ t: T) {
43         arr.append(t)
44     }
45     
46     func pop() -> T {
47         return arr.removeLast()
48     }
49 }

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