Chapter 09 C++ codes added.

Author: liuyubobobo

09-Dynamic-Programming/Course Code (C++)/06-0-1-knapsack-optimized/main.cpp

@@ -4,43 +4,48 @@
 
 using namespace std;
 
+/// 背包问题
+/// 动态规划改进: 滚动数组
+/// 时间复杂度: O(n * C) 其中n为物品个数; C为背包容积
+/// 空间复杂度: O(C), 实际使用了2*C的额外空间
 class Knapsack01{
 
 public:
     int knapsack01(const vector<int> &w, const vector<int> &v, int C){
-        assert( w.size() == v.size() && C >= 0 );
+        assert(w.size() == v.size() && C >= 0);
         int n = w.size();
         if( n == 0 && C == 0 )
             return 0;
 
-        vector<vector<int>> memo( 2, vector<int>(C+1,0));
+        vector<vector<int>> memo(2, vector<int>(C + 1, 0));
 
-        for( int j = 0 ; j <= C ; j ++ )
-            memo[0][j] = ( j >= w[0] ? v[0] : 0 );
+        for(int j = 0 ; j <= C ; j ++)
+            memo[0][j] = (j >= w[0] ? v[0] : 0);
 
-        for( int i = 1 ; i < n ; i ++ )
-            for( int j = 0 ; j <= C ; j ++ ){
-                memo[i%2][j] = memo[(i-1)%2][j];
-                if( j >= w[i] )
-                    memo[i%2][j] = max( memo[i%2][j], v[i] + memo[(i-1)%2][j-w[i]]);
+        for(int i = 1 ; i < n ; i ++)
+            for(int j = 0 ; j <= C ; j ++){
+                memo[i % 2][j] = memo[(i-1) % 2][j];
+                if(j >= w[i])
+                    memo[i % 2][j] = max(memo[i % 2][j], v[i] + memo[(i-1) % 2][j - w[i]]);
             }
-        return memo[(n-1)%2][C];
+        return memo[(n-1) % 2][C];
     }
 };
 
 int main() {
 
     int n, W;
-    cin>>n>>W;
+    cin >> n >> W;
 
-    int v,w;
+    int v, w;
     vector<int> vs, ws;
-    for( int i = 0 ; i < n ; i ++ ){
-        cin>>w>>v;
+    for(int i = 0 ; i < n ; i ++){
+        cin >> w >> v;
         vs.push_back(v);
         ws.push_back(w);
     }
 
-    cout<<Knapsack01().knapsack01(ws,vs,W)<<endl;
+    cout << Knapsack01().knapsack01(ws, vs, W) << endl;
+
     return 0;
 }

09-Dynamic-Programming/Course Code (C++)/06-0-1-knapsack-optimized/main2.cpp

@@ -4,23 +4,27 @@
 
 using namespace std;
 
+/// 背包问题
+/// 动态规划改进
+/// 时间复杂度: O(n * C) 其中n为物品个数; C为背包容积
+/// 空间复杂度: O(C), 只使用了C的额外空间
 class Knapsack01{
 
 public:
     int knapsack01(const vector<int> &w, const vector<int> &v, int C){
-        assert( w.size() == v.size() && C >= 0 );
-        int n = w.size();
-        if( n == 0 || C == 0 )
+        assert(w.size() == v.size() && C >= 0);
+        intn = w.size();
+        if(n == 0 || C == 0)
             return 0;
 
         vector<int> memo(C+1,0);
 
-        for( int j = 0 ; j <= C ; j ++ )
-            memo[j] = ( j >= w[0] ? v[0] : 0 );
+        for(int j = 0 ; j <= C ; j ++)
+            memo[j] = (j >= w[0] ? v[0] : 0);
 
-        for( int i = 1 ; i < n ; i ++ )
-            for( int j = C ; j >= w[i] ; j -- )
-                memo[j] = max( memo[j], v[i] + memo[j-w[i]]);
+        for(int i = 1 ; i < n ; i ++)
+            for(int j = C ; j >= w[i] ; j --)
+                memo[j] = max(memo[j], v[i] + memo[j - w[i]]);
 
         return memo[C];
     }
@@ -29,16 +33,17 @@ class Knapsack01{
 int main() {
 
     int n, W;
-    cin>>n>>W;
+    cin >> n >> W;
 
-    int v,w;
+    int v, w;
     vector<int> vs, ws;
-    for( int i = 0 ; i < n ; i ++ ){
-        cin>>w>>v;
+    for(int i = 0 ; i < n ; i ++){
+        cin >> w >> v;
         vs.push_back(v);
         ws.push_back(w);
     }
 
-    cout<<Knapsack01().knapsack01(ws,vs,W)<<endl;
+    cout << Knapsack01().knapsack01(ws, vs, W) << endl;
+
     return 0;
 }

09-Dynamic-Programming/Course Code (Java)/06-0-1-knapsack-optimized/src/Solution1.java

@@ -0,0 +1,37 @@
+/// 背包问题
+/// 动态规划改进: 滚动数组
+/// 时间复杂度: O(n * C) 其中n为物品个数; C为背包容积
+/// 空间复杂度: O(C), 实际使用了2*C的额外空间
+public class Solution1 {
+
+    public int knapsack01(int[] w, int[] v, int C){
+
+        if(w == null || v == null || w.length != v.length)
+            throw new IllegalArgumentException("Invalid w or v");
+
+        if(C < 0)
+            throw new IllegalArgumentException("C must be greater or equal to zero.");
+
+        int n = w.length;
+        if(n == 0 || C == 0)
+            return 0;
+
+        int[][] memo = new int[2][C + 1];
+
+        for(int j = 0 ; j <= C ; j ++)
+            memo[0][j] = (j >= w[0] ? v[0] : 0);
+
+        for(int i = 1 ; i < n ; i ++)
+            for(int j = 0 ; j <= C ; j ++){
+                memo[i % 2][j] = memo[(i-1) % 2][j];
+                if(j >= w[i])
+                    memo[i % 2][j] = Math.max(memo[i % 2][j], v[i] + memo[(i-1) % 2][j - w[i]]);
+            }
+
+        return memo[(n-1) % 2][C];
+    }
+
+    public static void main(String[] args) {
+
+    }
+}

09-Dynamic-Programming/Course Code (Java)/06-0-1-knapsack-optimized/src/Solution2.java

@@ -0,0 +1,34 @@
+/// 背包问题
+/// 动态规划改进
+/// 时间复杂度: O(n * C) 其中n为物品个数; C为背包容积
+/// 空间复杂度: O(C), 只使用了C的额外空间
+public class Solution2 {
+
+    public int knapsack01(int[] w, int[] v, int C){
+
+        if(w == null || v == null || w.length != v.length)
+            throw new IllegalArgumentException("Invalid w or v");
+
+        if(C < 0)
+            throw new IllegalArgumentException("C must be greater or equal to zero.");
+
+        int n = w.length;
+        if(n == 0 || C == 0)
+            return 0;
+
+        int[] memo = new int[C+1];
+
+        for(int j = 0 ; j <= C ; j ++)
+            memo[j] = (j >= w[0] ? v[0] : 0);
+
+        for(int i = 1 ; i < n ; i ++)
+            for(int j = C ; j >= w[i] ; j --)
+                memo[j] = Math.max(memo[j], v[i] + memo[j - w[i]]);
+
+        return memo[C];
+    }
+
+    public static void main(String[] args) {
+
+    }
+}