[Swift]LeetCode11. 盛最多水的容器 | Container With Most Water

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

给定 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0)。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。

说明：你不能倾斜容器，且 n 的值至少为 2。

图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下，容器能够容纳水（表示为蓝色部分）的最大值为 49。

双指针法

这种方法背后的思路在于，两线段之间形成的区域总是会受到其中较短那条长度的限制。此外，两线段距离越远，得到的面积就越大。

我们在由线段长度构成的数组中使用两个指针，一个放在开始，一个置于末尾。此外，我们会使用变量

查看下面的例子将有助于你更好地理解该算法：1 8 6 2 5 4 8 3 7

这种方法如何工作？

最初我们考虑由最外围两条线段构成的区域。现在，为了使面积最大化，我们需要考虑更长的两条线段之间的区域。如果我们试图将指向较长线段的指针向内侧移动，矩形区域的面积将受限于较短的线段而不会获得任何增加。但是，在同样的条件下，移动指向较短线段的指针尽管造成了矩形宽度的减小，但却可能会有助于面积的增大。因为移动较短线段的指针会得到一条相对较长的线段，这可以克服由宽度减小而引起的面积减小。

120ms

 1 class Solution {
 2     func maxArea(_ height: [Int]) -> Int {
 3         var maxarea:Int = 0, l:Int = 0, r:Int = height.count - 1
 4         while(l < r)
 5         {
 6             maxarea = max(maxarea, min(height[l], height[r]) * (r - l))
 7             if height[l] < height[r]
 8             {
 9                 l += 1
10             }  
11             else
12             {
13                 r -= 1
14             }   
15         }
16         return maxarea
17     }
18 }

20ms

 1 class Solution {
 2     func maxArea(_ height: [Int]) -> Int {
 3         var left = 0
 4         var right = height.count - 1
 5         var maxArea = area(height, left, right)
 6         while(left < right) {
 7             if (height[left] < height[right]) {
 8                 left = left + 1
 9             } else {
10                 right = right - 1
11             }
12             let lArea = area(height, left, right)
13             if (lArea > maxArea) {
14                 maxArea = lArea
15             }
16         }
17     return maxArea
18     }
19     
20     func area(_ height: [Int], _ start: Int, _ end: Int) -> Int {
21         let x = height[start]
22         let y = height[end]
23         return (end - start) * (x > y ? y : x)
24     }
25     
26 }

20ms

 1 class Solution {
 2         func maxArea(_ height: [Int]) -> Int {
 3         var result = 0
 4         
 5         var left = 0;
 6         var right = height.count - 1;
 7         
 8         while left < right {
 9             var valleft = height[left]
10             var valright = height[right]
11             let area = (valleft > valright ? valright : valleft) * (right - left)
12             
13             if area > result {
14                 result = area
15             }
16             
17             if valleft > valright {
18                 right -= 1
19             } else {
20                 left += 1
21             }
22         }
23         return result
24     }
25 }

24ms

 1 class Solution {
 2     func maxArea(_ height: [Int]) -> Int {
 3         var ans = 0
 4         
 5         var i = 0
 6         var j = height.count - 1
 7         
 8         while i < j {
 9             let h1 = height[i]
10             let h2 = height[j]
11             
12             let area = abs(i - j) * min(h1, h2)
13             ans = max(ans, area)
14             
15             if h1 < h2 { i += 1 }
16             else { j -= 1 }
17         }
18         
19         return ans
20     }
21 }

28ms

 1 class Solution {
 2     func maxArea(_ height: [Int]) -> Int {
 3         var area = 0
 4         var left = 0
 5         var right = 0
 6         
 7         var i = 0
 8         var j = height.count - 1
 9         
10         while i < j {
11             if left <= right {
12                 let tmpArea = left * (j - i)
13                 area = area < tmpArea ? tmpArea : area
14                 if left < height[i] {
15                     left = height[i]
16                 } else {
17                     i += 1
18                 }
19             } else {
20                 let tmpArea = right * (j - i)
21                 area = area < tmpArea ? tmpArea : area
22                 if right < height[j] {
23                     right = height[j]
24                 } else {
25                     j -= 1
26                 }
27             }
28         }
29         return area
30     }
31 }

40ms

 1 class Solution {
 2     func maxArea(_ height: [Int]) -> Int {
 3         var result = 0
 4         var i = 0
 5         var j = height.count - 1
 6             
 7         while (i < j) {
 8             let minHeight = min(height[i], height[j])
 9             let area = minHeight * (j - i)
10             result = max(result, area)
11             while (height[i] <= minHeight && i < j) { i += 1 }
12             while (height[j] <= minHeight && i < j) { j -= 1 }
13             print(i, j)
14         }
15         return result
16     }
17 }

108ms

 1 class Solution {
 2     func maxArea(_ height: [Int]) -> Int {
 3         if height.count == 0 { return 0 }
 4         if height.count == 1 { return height[0] }
 5         var head = 0
 6         var tail = height.count - 1
 7         let sortFunc: (((Int, Int), (Int, Int)) -> Bool) = { (ele1, ele2) -> Bool in
 8             return ele1.1 < ele2.1
 9         }
10 
11         var maxArea = 0
12         for (idx, h) in height.enumerated().sorted(by: sortFunc) {
13             while h > height[head] { head += 1 }
14             while h > height[tail] { tail -= 1 }
15 
16             let dToHead = idx - head
17             let dToTail = tail - idx
18             let area = max( dToHead * h, dToTail * h )
19             if area > maxArea {
20                 maxArea = area
21             }
22         }
23 
24         return maxArea
25     }
26 }

[Swift]LeetCode11. 盛最多水的容器 | Container With Most Water

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