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Author: assassint2017

House Robber/House_Robber.cpp

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+// 思路一 简单的回溯超时了
+// 59 / 69 test cases passed.
+class Solution 
+{
+public:
+    int rob(vector<int>& nums) 
+    {
+        return helper(nums, 0);
+    }
+private:
+    int helper(vector<int>& nums, int start)
+    {
+        if (start >= nums.size())
+            return 0;
+        else if (start == nums.size() - 1)
+            return nums[start];
+        else
+            return max(nums[start] + helper(nums, start + 2), nums[start + 1] + helper(nums, start + 3));
+    }
+};
+
+// 思路二
+// 自底向上的动态规划
+// Runtime: 4 ms, faster than 100.00% of C++ online submissions for House Robber.
+// Memory Usage: 9.1 MB, less than 11.45% of C++ online submissions for House Robber.
+class Solution 
+{
+public:
+    int rob(vector<int>& nums) 
+    {
+        if (nums.size() == 0)
+            return 0;
+
+        int *res = new int[nums.size()];
+
+        // 边界初始化
+        res[nums.size() - 1] = nums[nums.size() - 1];
+        if (nums.size() >= 2)
+            res[nums.size() - 2] = max(nums[nums.size() - 1], nums[nums.size() - 2]);
+        if (nums.size() >= 3)
+            res[nums.size() - 3] = max(nums[nums.size() - 1] + nums[nums.size() - 3], nums[nums.size() - 2]);
+
+        for (int i = nums.size() - 4; i >= 0; i--)
+            res[i] = max(nums[i] + res[i + 2], nums[i + 1] + res[i + 3]);
+
+        int finres = res[0];
+        delete[] res;
+        return finres;
+    }
+};

House Robber/House_Robber.py

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+# Runtime: 44 ms, faster than 34.83% of Python3 online submissions for House Robber.
+# Memory Usage: 13.1 MB, less than 5.00% of Python3 online submissions for House Robber.
+
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) is 0:
+            return 0
+        
+        res = [0 for i in range(len(nums))]
+
+        res[-1] = nums[-1]
+        if len(nums) >= 2:
+            res[-2] = max(nums[-1], nums[-2])
+        if len(nums) >= 3:
+            res[-3] = max(nums[-1] + nums[-3], nums[-2])
+        
+        for i in range(len(nums) - 4, -1, -1):
+            res[i] = max(nums[i] + res[i + 2], nums[i + 1] + res[i + 3])
+        
+        return res[0]