leetcode刷题笔记十一 盛最多水的容器 Scala版本

leetcode刷题笔记十一 盛最多水的容器 Scala版本

源地址：盛最多水的容器

问题描述：

Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

简要思路分析：

首先想到，可以通过暴力法计算任意两条线与X轴构成的面积，通过变量记录其中的最大值，并返回。但是这种复杂度较高。

易知，最终面积是由连续的点构成的，我们可以视为最终面积是从起始点到终点构成的最初面积，不断舍去两侧的高度较低的线，使得面积更加集中。因此，我们可以通过双指针扫描法确定最大面积。

代码补充：

object Solution {
   def maxArea(height: Array[Int]): Int = {
      var maxArea = 0
      var leftArr = 0
      var rightArr = height.length-1

      while (leftArr < rightArr) {
        //println("-------------------------------------")
        //println("leftArr: "+ leftArr)
        //println("height(leftArr): "+height(leftArr))
        //println("rightArr:" + rightArr)
        //println("height(rightArr): "+height(rightArr))
        //println("maxArea: "+maxArea)
        var Area = (math.min(height(leftArr), height(rightArr)) * (rightArr-leftArr))
        //if (Area < maxArea) return maxArea
        //else {
          if (height(leftArr) < height(rightArr)) leftArr+=1
          else rightArr-=1
          if (Area> maxArea)maxArea = Area
        //}

      }
      return maxArea
    }
}

func maxArea(height []int) int {
    ans := 0
    for i , j := 0, len(height) -1; i < j;{
        ans = max(ans, min(height[i], height[j]) * (j - i))
        if height[i] > height[j] {
            j -= 1
        } else {
            i += 1
        }
    }
    return ans
}

func min(a int, b int) int {
    if a > b {
        return b
    } else {
        return a
    }
}

func max(a int, b int) int {
    if a > b {
        return a
    } else {
        return b
    }
}

//使用单调栈处理
func maxArea(height []int) int {
    var left = stack{}
    var right = stack{}
    res := 0
    left.Push(0)
    
    for i, v := range height {
        if height[left.Top()] < v {
            left.Push(i)
        }
        for (len(right.arr) != 0 && height[right.Top()] < height[i]) {
            right.Pop()
        }
        right.Push(i)
    }

    for (len(right.arr) > 0){
        r := right.Top()
        right.Pop()
        for i := 0; i < len(left.arr); i++{
            l := left.arr[i]
            if l > r { break }
            curHeight := min(height[l], height[r])
            res = max(res, curHeight * (r-l))
        }
    }

    return res

}

type stack struct{
    arr []int
}

func (s *stack) Push (val int){
    s.arr = append(s.arr, val)
}

func (s *stack) Pop() {
    var _ = s.arr[len(s.arr)-1]
    s.arr = s.arr[0:len(s.arr)-1]
    //return val
}

func (s *stack) Top() int{
    return s.arr[len(s.arr)-1] 
}

func max(a, b int) int {
    if a > b {
        return a
    } else {
        return b
    }
}

func min(a, b int) int {
    if a < b {
        return a
    } else {
        return b
    }
}