[Swift]LeetCode42.接雨水|TrappingRainWater

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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.

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!

Example:

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

给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。

上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。感谢 Marcos 贡献此图。

示例:

输入: [0,1,0,2,1,0,1,3,2,1,2,1]
输出: 6

【双指针】12ms

我们可能会想到在一次迭代中做某种方式，而不是单独计算左右部分。
从图中的动态编程方法来看，只要注意 $right_max 在迭代期间，但现在我们可以使用2个指针在一次迭代中完成，在两者之间切换。$

算法

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 $ans$
  - 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 }

【动态编程】 16ms

 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】28ms

 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 }

【使用堆栈】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 }