11. Container With Most Water

leetcode/0011.Container-With-Most-Water/11. Container With Most Water.go

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+package leetcode
+
+func ContainerWithMostWater(height []int) int {
+	left, right := 0, len(height)-1
+	maxCapacity := 0
+	for right > left {
+		capacity := min(height[left], height[right]) * (right - left)
+		if capacity > maxCapacity {
+			maxCapacity = capacity
+		}
+		if height[left] <= height[right] {
+			left++
+		} else {
+			right--
+		}
+	}
+	return maxCapacity
+}

leetcode/0011.Container-With-Most-Water/11. Container With Most Water_test.go

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+package leetcode
+
+import (
+	"reflect"
+	"testing"
+)
+
+// 填入 function input type
+type parameters0011 struct {
+	height []int
+}
+
+func TestContainerWithMostWater(t *testing.T) {
+	tests := []struct {
+		parameters0011
+		// 填入 function output type
+		ans int
+	}{
+		// 填入 test case
+		{
+			parameters0011{[]int{1, 8, 6, 2, 5, 4, 8, 3, 7}},
+			49,
+		},
+		{
+			parameters0011{[]int{1, 1}},
+			1,
+		},
+	}
+
+	for _, test := range tests {
+		t.Run("Test ContainerWithMostWater", func(t *testing.T) {
+			// 完整輸入參數
+			result := ContainerWithMostWater(test.parameters0011.height)
+			// compare 的方式需視情況調整
+			if !reflect.DeepEqual(result, test.ans) {
+				t.Errorf("ContainerWithMostWater(%+v) got %+v, want %+v", test.parameters0011, result, test.ans)
+			}
+		})
+	}
+}

leetcode/0011.Container-With-Most-Water/README.md

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+# 11. Container With Most Water
+
+<https://leetcode.com/problems/container-with-most-water/>
+
+![example 1](image1.png)
+
+給定數字陣列 `height`，找出能容納最多水容量的空間為多少  
+要找出使用的是哪兩個 `heigth`，容器的容積等於 $min(h1, h2) \times (x2-x1)$
+
+由於容器一定由左邊和右邊構成，所以可以使用雙指針，並且一個從左一個從右開始  
+計算出當下的容積後，移動高度較低的指針，往中間靠近  
+如此便能使用 $O(n)$ 找出最大的容積
+
+## Takeaway
+
+- Two Points