[Swift]LeetCode42. 接雨水 | Trapping Rain Water

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➤微信公众号：山青咏芝（shanqingyongzhi）
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➤原文地址：https://www.cnblogs.com/strengthen/p/9903130.html 
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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 [ i ] > left_max [ i ]right_max[一世]>left_max[一世] （从元素0到6），被困水取决于left_max，类似情况 left_max [ i ] > right_max [ i ]left_max[一世]>right_max[一世]（从要素8到11）。
因此，我们可以说如果一端有一个较大的条（如右图），我们可以确保被困的水将取决于当前方向（从左到右）的条形高度。
一旦我们发现另一端（右）的条形较小，我们就会开始以相反的方向（从右到左）进行迭代。
我们必须坚持left_maxleft_max 和 right_maxright_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  ext{left}left pointer to 0 and  ext{right}right pointer to size-1
• While  ext{left}&lt; ext{right}left<right, do:
  • If  ext{height[left]}height[left] is smaller than  ext{height[right]}height[right]
  • If height[left]>=left_maxheight[left]>=left_max, update left_maxleft_max
    • Else add left_max−height[left]left_max−height[left] to  ext{ans}ans
    • Add 1 to  ext{left}left.
  • Else
  • If height[right]>=right_maxheight[right]>=right_max, update right_maxright_max
  • Else add right_max−height[right]right_max−height[right] to  ext{ans}ans
  • Subtract 1 from  ext{right}right.

class Solution {
    func trap(_ height: [Int]) -> Int {
        var leftMax = 0, rightMax = 0
        var leftPointer = 0, rightPointer = height.count - 1
        var trappedWater = 0
        
        while leftPointer < rightPointer {
            if height[leftPointer] < height[rightPointer] {
                if height[leftPointer] > leftMax {
                    leftMax = height[leftPointer]
                } else {
                    trappedWater += leftMax - height[leftPointer]
                }
                leftPointer += 1
            } else {
                 if height[rightPointer] > rightMax {
                    rightMax = height[rightPointer]
                } else {
                    trappedWater += rightMax - height[rightPointer]
                }
                rightPointer -= 1
            } 
        }
        return trappedWater
    }
}

---

【动态编程】 16ms

class Solution {
  func trap(_ height: [Int]) -> Int {
    var leftMax = [Int]()
    var rightMax = [Int]()
    var maxL = 0
    var maxR = 0
    for i in 0..<height.count {
      if maxL < height[i] {
        maxL = height[i]
      }
      leftMax.append(maxL)
      if maxR < height[height.count - 1 - i] {
        maxR = height[height.count - 1 - i]
      }
      rightMax.append(maxR)
    }
    rightMax.reverse()
    var result = 0
    for i in 0..<height.count {
      let wall = max(min(leftMax[i], rightMax[i]), height[i])
      result += wall - height[i]
    }
    return result
  }
}

---

【暴力破解】28ms

class Solution {
    func trap(_ height: [Int]) -> Int {
        if height.count <= 0 {
            return 0
        }
        var maxL = height[0]
        var rights : Array = Array<Int>(repeating: 0, count: height.count)
        var result = 0
        var maxR = 0
        
        for i in height.enumerated().reversed() {
            if height[i.offset] > maxR {
                maxR = i.element
                rights[i.offset] = maxR
            }else {
                rights[i.offset] = maxR
            }
        }
        
        for i in 0..<height.count {
            if height[i] > maxL {
                maxL = height[i]
            }
            result += max(min(maxL, rights[i]) - height[i],0)
        }
        
        
        return result
    }
}

---

【暴力破解】44ms

class Solution {
    func trap(_ height: [Int]) -> Int {
        guard height.count > 2 else { return 0 }
        var res: Int = 0
        var leftMax = Array(repeating: 0, count: height.count)
        var rightMax = Array(repeating: 0, count: height.count)
    
        leftMax[0] = height[0]
        for i in 1..<height.count {
            leftMax[i] = max(leftMax[i - 1], height[i])
        }
        
        rightMax[height.count - 1] = height[height.count - 1]
        for i in (0..<height.count - 1).reversed() {
            rightMax[i] = max(rightMax[i + 1], height[i])
        }
    
        for i in 1..<height.count - 1 {
            res += min(leftMax[i], rightMax[i]) - height[i]
        }
        
        return res
    }
}

---

 【使用堆栈】64ms

class Solution {
    func trap(_ height: [Int]) -> Int {
        var stack = Stack<Int>()
        
        var ans = 0
        
        var current = 0
        
        while current < height.count {
            while !stack.isEmpty && height[current] > height[stack.peek()] {
                let top = stack.pop()

                if stack.isEmpty {
                    break
                }
                
                let distance = current - stack.peek() - 1
                
                let height = min(height[current], height[stack.peek()]) - height[top]
                
                ans += distance * height
            }
            stack.push(current)
            current += 1
        }
        
        return ans
    }
}

class Stack<T> {
    private var arr = [T]()
    
    var isEmpty: Bool {
        return arr.isEmpty
    }
    
    func peek() -> T {
        return arr.last!
    }
    
    func push(_ t: T) {
        arr.append(t)
    }
    
    func pop() -> T {
        return arr.removeLast()
    }
}