Chapter 09 section 04 codes C++ codes updated. Java codes added. new optimized algo added.

Author: liuyubobobo

09-Dynamic-Programming/Course Code (C++)/04-House-Robber/CMakeLists.txt

@@ -3,5 +3,5 @@ project(04_House_Robber)
 
 set(CMAKE_CXX_FLAGS "${CMAKE_CXX_FLAGS} -std=c++11")
 
-set(SOURCE_FILES main4.cpp)
+set(SOURCE_FILES main5.cpp)
 add_executable(04_House_Robber ${SOURCE_FILES})

09-Dynamic-Programming/Course Code (C++)/04-House-Robber/main.cpp

@@ -3,35 +3,49 @@
 
 using namespace std;
 
+/// 198. House Robber
+/// https://leetcode.com/problems/house-robber/description/
+/// 记忆化搜索
+/// 时间复杂度: O(n^2)
+/// 空间复杂度: O(n)
 class Solution {
+
 private:
     // memo[i] 表示考虑抢劫 nums[i...n) 所能获得的最大收益
     vector<int> memo;
 
     // 考虑抢劫nums[index...nums.size())这个范围的所有房子
-    int tryRob( vector<int> &nums, int index){
+    int tryRob(const vector<int> &nums, int index){
 
-        if( index >= nums.size() )
+        if(index >= nums.size())
             return 0;
 
-        if( memo[index] != -1 )
+        if(memo[index] != -1)
             return memo[index];
 
         int res = 0;
-        for( int i = index ; i < nums.size() ; i ++ )
-            res = max(res, nums[i] + tryRob(nums, i+2));
+        for(int i = index ; i < nums.size() ; i ++)
+            res = max(res, nums[i] + tryRob(nums, i + 2));
         memo[index] = res;
         return res;
     }
+
 public:
     int rob(vector<int>& nums) {
 
-        memo = vector<int>(nums.size(), -1);
+        memo.clear();
+        for(int i = 0 ; i < nums.size() ; i ++)
+            memo.push_back(-1);
         return tryRob(nums, 0);
     }
 };
 
 int main() {
 
+    int nums[] = {2, 1};
+    vector<int> vec(nums, nums + sizeof(nums)/sizeof(int));
+
+    cout << Solution().rob(vec) << endl;
+
     return 0;
 }

09-Dynamic-Programming/Course Code (C++)/04-House-Robber/main2.cpp

@@ -3,22 +3,28 @@
 
 using namespace std;
 
+/// 198. House Robber
+/// https://leetcode.com/problems/house-robber/description/
+/// 动态规划
+/// 时间复杂度: O(n^2)
+/// 空间复杂度: O(n)
 class Solution {
 
 public:
     int rob(vector<int>& nums) {
 
         int n = nums.size();
 
-        if( n == 0 )
+        if(n == 0)
             return 0;
 
         // memo[i] 表示考虑抢劫 nums[i...n) 所能获得的最大收益
         vector<int> memo(n, 0);
-        memo[n-1] = nums[n-1];
-        for( int i = n-2 ; i >= 0 ; i -- )
+        memo[n - 1] = nums[n - 1];
+        for(int i = n - 2 ; i >= 0 ; i --)
             for (int j = i; j < n; j++)
-                memo[i] = max(memo[i], nums[j] + (j + 2 < n ? memo[j + 2] : 0) );
+                memo[i] = max(memo[i],
+                              nums[j] + (j + 2 < n ? memo[j + 2] : 0));
 
         return memo[0];
     }
@@ -27,9 +33,9 @@ class Solution {
 int main() {
 
     int nums[] = {2,1};
-    vector<int> vec(nums, nums+sizeof(nums)/sizeof(int));
+    vector<int> vec(nums, nums + sizeof(nums)/sizeof(int));
 
-    cout<<Solution().rob(vec)<<endl;
+    cout << Solution().rob(vec) << endl;
 
     return 0;
 }

09-Dynamic-Programming/Course Code (C++)/04-House-Robber/main3.cpp

@@ -3,36 +3,49 @@
 
 using namespace std;
 
-// 改变状态定义
+/// 198. House Robber
+/// https://leetcode.com/problems/house-robber/description/
+/// 记忆化搜索, 改变状态定义
+/// 时间复杂度: O(n^2)
+/// 空间复杂度: O(n)
 class Solution {
+
 private:
     // memo[i] 表示考虑抢劫 nums[0...i] 所能获得的最大收益
     vector<int> memo;
 
     // 考虑抢劫nums[0...index]这个范围的所有房子
-    int tryRob( vector<int> &nums, int index){
+    int tryRob(const vector<int> &nums, int index){
 
-        if( index < 0 )
+        if(index < 0)
             return 0;
 
-        if( memo[index] != -1 )
+        if(memo[index] != -1)
             return memo[index];
 
         int res = 0;
-        for( int i = 0 ; i <= index ; i ++ )
-            res = max(res, nums[i] + tryRob(nums, i-2));
+        for(int i = 0 ; i <= index ; i ++)
+            res = max(res, nums[i] + tryRob(nums, i - 2));
         memo[index] = res;
         return res;
     }
+
 public:
     int rob(vector<int>& nums) {
 
-        memo = vector<int>(nums.size(), -1);
+        memo.clear();
+        for(int i = 0 ; i < nums.size() ; i ++)
+            memo.push_back(-1);
         return tryRob(nums, nums.size() - 1 );
     }
 };
 
 int main() {
 
+    int nums[] = {2, 1};
+    vector<int> vec(nums, nums + sizeof(nums)/sizeof(int));
+
+    cout << Solution().rob(vec) << endl;
+
     return 0;
 }

09-Dynamic-Programming/Course Code (C++)/04-House-Robber/main4.cpp

@@ -3,7 +3,11 @@
 
 using namespace std;
 
-// 改变状态定义
+/// 198. House Robber
+/// https://leetcode.com/problems/house-robber/description/
+/// 动态规划, 改变状态定义
+/// 时间复杂度: O(n^2)
+/// 空间复杂度: O(n)
 class Solution {
 
 public:
@@ -16,20 +20,21 @@ class Solution {
         // memo[i] 表示考虑抢劫 nums[0...i] 所能获得的最大收益
         vector<int> memo(n, 0);
         memo[0] = nums[0];
-        for( int i = 1 ; i < n ; i ++ )
+        for(int i = 1 ; i < n ; i ++)
             for (int j = i; j >= 0; j--)
-                memo[i] = max(memo[i], nums[j] + (j - 2 >= 0 ? memo[j - 2] : 0) );
+                memo[i] = max(memo[i],
+                              nums[j] + (j - 2 >= 0 ? memo[j - 2] : 0));
 
         return memo[n-1];
     }
 };
 
 int main() {
 
-    int nums[] = {2,1};
-    vector<int> vec(nums, nums+sizeof(nums)/sizeof(int));
+    int nums[] = {2, 1};
+    vector<int> vec(nums, nums + sizeof(nums)/sizeof(int));
 
-    cout<<Solution().rob(vec)<<endl;
+    cout << Solution().rob(vec) << endl;
 
     return 0;
 }

09-Dynamic-Programming/Course Code (C++)/04-House-Robber/main5.cpp

@@ -0,0 +1,49 @@
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+/// 198. House Robber
+/// https://leetcode.com/problems/house-robber/description/
+/// 记忆化搜索, 优化状态转移
+/// 时间复杂度: O(n)
+/// 空间复杂度: O(n)
+class Solution {
+private:
+    // memo[i] 表示考虑抢劫 nums[i...n) 所能获得的最大收益
+    vector<int> memo;
+
+    // 考虑抢劫nums[index...nums.size())这个范围的所有房子
+    int tryRob(const vector<int> &nums, int index){
+
+        if(index >= nums.size())
+            return 0;
+
+        if(memo[index] != -1)
+            return memo[index];
+
+        // 或者当前房子放弃, 从下一个房子开始考虑
+        // 或者抢劫当前的房子, 从i+2以后的房子开始考虑
+        return memo[index] = max(tryRob(nums, index + 1),
+                                 nums[index] + tryRob(nums, index + 2));
+    }
+
+public:
+    int rob(vector<int>& nums) {
+
+        memo.clear();
+        for(int i = 0 ; i < nums.size() ; i ++)
+            memo.push_back(-1);
+        return tryRob(nums, 0);
+    }
+};
+
+int main() {
+
+    int nums[] = {2, 1};
+    vector<int> vec(nums, nums + sizeof(nums)/sizeof(int));
+
+    cout << Solution().rob(vec) << endl;
+
+    return 0;
+}

09-Dynamic-Programming/Course Code (C++)/04-House-Robber/main6.cpp

@@ -0,0 +1,42 @@
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+/// 198. House Robber
+/// https://leetcode.com/problems/house-robber/description/
+/// 动态规划, 优化状态转移
+/// 时间复杂度: O(n)
+/// 空间复杂度: O(n)
+class Solution {
+
+public:
+    int rob(vector<int>& nums) {
+
+        int n = nums.size();
+
+        if(n == 0)
+            return 0;
+
+        // memo[i] 表示考虑抢劫 nums[i...n) 所能获得的最大收益
+        vector<int> memo(n, 0);
+        memo[n - 1] = nums[n - 1];
+        for(int i = n - 2 ; i >= 0 ; i --)
+            // 或者当前房子放弃, 从下一个房子开始考虑
+            // 或者抢劫当前的房子, 从i+2以后的房子开始考虑
+            memo[i] = max(memo[i + 1],
+                          nums[i] + (i + 2 < n ? memo[i + 2] : 0));
+
+        return memo[0];
+    }
+};
+
+int main() {
+
+    int nums[] = {2, 1};
+    vector<int> vec(nums, nums + sizeof(nums)/sizeof(int));
+
+    cout << Solution().rob(vec) << endl;
+
+    return 0;
+}

09-Dynamic-Programming/Course Code (C++)/04-House-Robber/main7.cpp

@@ -0,0 +1,50 @@
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+/// 198. House Robber
+/// https://leetcode.com/problems/house-robber/description/
+/// 记忆化搜索, 改变状态定义, 优化转移方程
+/// 时间复杂度: O(n)
+/// 空间复杂度: O(n)
+class Solution {
+
+private:
+    // memo[i] 表示考虑抢劫 nums[0...i] 所能获得的最大收益
+    vector<int> memo;
+
+    // 考虑抢劫nums[0...index]这个范围的所有房子
+    int tryRob(const vector<int> &nums, int index){
+
+        if(index < 0)
+            return 0;
+
+        if(memo[index] != -1)
+            return memo[index];
+
+        // 或者当前房子放弃, 考虑[0...index-1]的所有房子
+        // 或者抢劫当前的房子, 考虑[0...index-2]的所有房子
+        return memo[index] = max(tryRob(nums, index - 1),
+                                 nums[index] + tryRob(nums, index - 2));
+    }
+
+public:
+    int rob(vector<int>& nums) {
+
+        memo.clear();
+        for(int i = 0 ; i < nums.size() ; i ++)
+            memo.push_back(-1);
+        return tryRob(nums, nums.size() - 1 );
+    }
+};
+
+int main() {
+
+    int nums[] = {2, 1};
+    vector<int> vec(nums, nums + sizeof(nums)/sizeof(int));
+
+    cout << Solution().rob(vec) << endl;
+
+    return 0;
+}

09-Dynamic-Programming/Course Code (C++)/04-House-Robber/main8.cpp

@@ -0,0 +1,39 @@
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+/// 198. House Robber
+/// https://leetcode.com/problems/house-robber/description/
+/// 动态规划, 改变状态定义, 优化转移方程
+/// 时间复杂度: O(n)
+/// 空间复杂度: O(n)
+class Solution {
+
+public:
+    int rob(vector<int>& nums) {
+
+        int n = nums.size();
+        if( n == 0 )
+            return 0;
+
+        // memo[i] 表示考虑抢劫 nums[0...i] 所能获得的最大收益
+        vector<int> memo(n, 0);
+        memo[0] = nums[0];
+        for(int i = 1 ; i < n ; i ++)
+            memo[i] = max(memo[i - 1],
+                          nums[i] + (i - 2 >= 0 ? memo[i - 2] : 0));
+
+        return memo[n-1];
+    }
+};
+
+int main() {
+
+    int nums[] = {2, 1};
+    vector<int> vec(nums, nums + sizeof(nums)/sizeof(int));
+
+    cout << Solution().rob(vec) << endl;
+
+    return 0;
+}