01背包-2增加Go一维数组,二维数组移至-1

problems/背包理论基础01背包-1.md

@@ -437,6 +437,58 @@ print(dp[n - 1][bagweight])
 ### Go
 
 ```go
+package main
+
+import (
+	"fmt"
+)
+
+func main() {
+	var n, bagweight int // bagweight代表行李箱空间
+	fmt.Scan(&n, &bagweight)
+
+	weight := make([]int, n) // 存储每件物品所占空间
+	value := make([]int, n)  // 存储每件物品价值
+
+	for i := 0; i < n; i++ {
+		fmt.Scan(&weight[i])
+	}
+	for j := 0; j < n; j++ {
+		fmt.Scan(&value[j])
+	}
+	// dp数组, dp[i][j]代表行李箱空间为j的情况下,从下标为[0, i]的物品里面任意取,能达到的最大价值
+	dp := make([][]int, n)
+	for i := range dp {
+		dp[i] = make([]int, bagweight+1)
+	}
+
+	// 初始化, 因为需要用到dp[i - 1]的值
+	// j < weight[0]已在上方被初始化为0
+	// j >= weight[0]的值就初始化为value[0]
+	for j := weight[0]; j <= bagweight; j++ {
+		dp[0][j] = value[0]
+	}
+
+	for i := 1; i < n; i++ { // 遍历科研物品
+		for j := 0; j <= bagweight; j++ { // 遍历行李箱容量
+			if j < weight[i] {
+				dp[i][j] = dp[i-1][j] // 如果装不下这个物品,那么就继承dp[i - 1][j]的值
+			} else {
+				dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
+			}
+		}
+	}
+
+	fmt.Println(dp[n-1][bagweight])
+}
+
+func max(x, y int) int {
+	if x > y {
+		return x
+	}
+	return y
+}
+
 ```
 
 ### Javascript

problems/背包理论基础01背包-2.md

@@ -318,50 +318,45 @@ print(dp[n - 1][bagweight])
 package main
 
 import (
-    "fmt"
+	"fmt"
 )
 
 func main() {
-    var n, bagweight int
-    fmt.Scan(&n, &bagweight)
+	// 读取 M 和 N
+	var M, N int
+	fmt.Scan(&M, &N)
 
-    weight := make([]int, n)
-    value := make([]int, n)
+	costs := make([]int, M)
+	values := make([]int, M)
 
-    for i := 0; i < n; i++ {
-        fmt.Scan(&weight[i])
-    }
-    for j := 0; j < n; j++ {
-        fmt.Scan(&value[j])
-    }
+	for i := 0; i < M; i++ {
+		fmt.Scan(&costs[i])
+	}
+	for j := 0; j < M; j++ {
+		fmt.Scan(&values[j])
+	}
 
-    dp := make([][]int, n)
-    for i := range dp {
-        dp[i] = make([]int, bagweight+1)
-    }
+	// 创建一个动态规划数组dp，初始值为0
+	dp := make([]int, N + 1)
 
-    for j := weight[0]; j <= bagweight; j++ {
-        dp[0][j] = value[0]
-    }
+	// 外层循环遍历每个类型的研究材料
+	for i := 0; i < M; i++ {
+		// 内层循环从 N 空间逐渐减少到当前研究材料所占空间
+		for j := N; j >= costs[i]; j-- {
+			// 考虑当前研究材料选择和不选择的情况，选择最大值
+			dp[j] = max(dp[j], dp[j-costs[i]] + values[i])
+		}
+	}
 
-    for i := 1; i < n; i++ {
-        for j := 0; j <= bagweight; j++ {
-            if j < weight[i] {
-                dp[i][j] = dp[i-1][j]
-            } else {
-                dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
-            }
-        }
-    }
-
-    fmt.Println(dp[n-1][bagweight])
+	// 输出dp[N]，即在给定 N 行李空间可以携带的研究材料最大价值
+	fmt.Println(dp[N])
 }
 
-func max(a, b int) int {
-    if a > b {
-        return a
-    }
-    return b
+func max(x, y int) int {
+	if x > y {
+		return x
+	}
+	return y
 }
 
 ```