Chapter 09 section 05 C++ codes updated. Java codes added.

Author: liuyubobobo

09-Dynamic-Programming/Course Code (C++)/05-0-1-knapsack/main.cpp

@@ -4,51 +4,59 @@
 
 using namespace std;
 
+/// 背包问题
+/// 记忆化搜索
+/// 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(n * C)
 class Knapsack01{
 
 private:
     vector<vector<int>> memo;
 
     // 用 [0...index]的物品,填充容积为c的背包的最大价值
-    int bestValue(const vector<int> &w, const vector<int> v, int index, int c){
+    int bestValue(const vector<int> &w, const vector<int> &v, int index, int c){
 
-        if( c <= 0 || index < 0 )
+        if(c <= 0 || index < 0)
             return 0;
 
-        if( memo[index][c] != -1 )
+        if(memo[index][c] != -1)
             return memo[index][c];
 
         int res = bestValue(w, v, index-1, c);
-        if( c >= w[index] )
-            res = max( res , v[index] + bestValue(w, v, index-1, c-w[index]) );
+        if(c >= w[index])
+            res = max(res, v[index] + bestValue(w, v, index - 1, c - w[index]));
         memo[index][c] = res;
         return res;
     }
+
 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 )
+        if(n == 0 || C == 0)
             return 0;
 
-        memo = vector<vector<int>>( n, vector<int>(C+1,-1));
-        return bestValue(w, v, n-1, C);
+        memo.clear();
+        for(int i = 0 ; i < n ; i ++)
+            memo.push_back(vector<int>(C + 1, -1));
+        return bestValue(w, v, n - 1, 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++)/05-0-1-knapsack/main2.cpp

@@ -4,43 +4,48 @@
 
 using namespace std;
 
+/// 背包问题
+/// 动态规划
+/// 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(n * C)
 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 )
+        if(n == 0 || C == 0)
             return 0;
 
-        vector<vector<int>> memo( n, vector<int>(C+1,0));
+        vector<vector<int>> memo(n, 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 ++ ){
+        for(int i = 1 ; i < n ; i ++)
+            for(int j = 0 ; j <= C ; j ++){
                 memo[i][j] = memo[i-1][j];
-                if( j >= w[i] )
-                    memo[i][j] = max( memo[i][j], v[i] + memo[i-1][j-w[i]]);
+                if(j >= w[i])
+                    memo[i][j] = max(memo[i][j], v[i] + memo[i - 1][j - w[i]]);
             }
-        return memo[n-1][C];
+        return memo[n - 1][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 (Java)/05-0-1-knapsack/src/Solution1.java

@@ -0,0 +1,45 @@
+/// 背包问题
+/// 记忆化搜索
+/// 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(n * C)
+public class Solution1 {
+
+    private int[][] memo;
+
+    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;
+
+        memo = new int[n][C + 1];
+        return bestValue(w, v, n - 1, C);
+    }
+
+    // 用 [0...index]的物品,填充容积为c的背包的最大价值
+    private int bestValue(int[] w, int[] v, int index, int c){
+
+        if(c <= 0 || index < 0)
+            return 0;
+
+        if(memo[index][c] != -1)
+            return memo[index][c];
+
+        int res = bestValue(w, v, index-1, c);
+        if(c >= w[index])
+            res = Math.max(res, v[index] + bestValue(w, v, index - 1, c - w[index]));
+
+        return memo[index][c] = res;
+    }
+
+    public static void main(String[] args) {
+
+    }
+
+}

09-Dynamic-Programming/Course Code (Java)/05-0-1-knapsack/src/Solution2.java

@@ -0,0 +1,37 @@
+/// 背包问题
+/// 动态规划
+/// 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(n * C)
+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[n][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][j] = memo[i-1][j];
+                if(j >= w[i])
+                    memo[i][j] = Math.max(memo[i][j], v[i] + memo[i - 1][j - w[i]]);
+            }
+
+        return memo[n - 1][C];
+    }
+
+    public static void main(String[] args) {
+
+    }
+}