0198-house-robber.md Added 7 languages' solutions.

gazeldx · gazeldx · commit 7c2a8b32715f · 2024-12-05T11:20:12.000+08:00
diff --git a/README.md b/README.md
@@ -19,6 +19,8 @@ Generally speaking, the simple questions are placed at the front, and the depend
 After finishing one category of questions, you can study another category to improve your sense of achievement and learning speed.
 
 ## Dynamic programming
+- [198. House Robber](problems/0198-house-robber.md)
+
 - [53. Maximum Subarray](problems/0053-maximum-subarray.md)
 - [392. Is Subsequence](problems/0392-is-subsequence.md)
 - [583. Delete Operation for Two Strings](problems/0583-delete-operation-for-two-strings.md)
diff --git a/problems/0198-house-robber.md b/problems/0198-house-robber.md
@@ -0,0 +1,171 @@
+# 198. House Robber
+LeetCode problem: [198. House Robber](https://leetcode.com/problems/house-robber/)
+
+## LeetCode problem description
+> You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.
+
+Given an integer array `nums` representing the amount of money of each house, return the maximum amount of money you can rob tonight **without alerting the police**.
+```
+-------------------------------------------------------------------------------------------------
+[Example 1]
+
+Input: nums = [1,2,3,1]
+Output: 4
+Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
+             Total amount you can rob = 1 + 3 = 4.
+-------------------------------------------------------------------------------------------------
+[Example 2]
+Input: nums = [2,7,9,3,1]
+Output: 12
+Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
+             Total amount you can rob = 2 + 9 + 1 = 12.
+-------------------------------------------------------------------------------------------------
+[Constraints]
+
+1 <= nums.length <= 100
+0 <= nums[i] <= 400
+-------------------------------------------------------------------------------------------------
+```
+
+## Thoughts
+This problem can be solved using **Dynamic programming**.
+
+### Complexity
+* Time: `O(n)`.
+* Space: `O(n)`.
+
+## Python
+```python
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) == 1:
+            return nums[0]
+
+        dp = [0] * len(nums)
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+
+        for i in range(2, len(dp)):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+
+        return dp[-1]
+```
+
+## C++
+```cpp
+class Solution {
+public:
+    int rob(vector<int>& nums) {
+        if (nums.size() == 1) {
+            return nums[0];
+        }
+
+        auto dp = vector<int>(nums.size());
+        dp[0] = nums[0];
+        dp[1] = max(nums[0], nums[1]);
+
+        for (auto i = 2; i < dp.size(); i++) {
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]);
+        }
+
+        return dp.back();
+    }
+};
+```
+
+## Java
+```java
+class Solution {
+    public int rob(int[] nums) {
+        if (nums.length == 1) {
+            return nums[0];
+        }
+
+        var dp = new int[nums.length];
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+
+        for (var i = 2; i < dp.length; i++) {
+            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
+        }
+
+        return dp[dp.length - 1];
+    }
+}
+```
+
+## C#
+```c#
+public class Solution {
+    public int Rob(int[] nums) {
+        if (nums.Length == 1) {
+            return nums[0];
+        }
+
+        var dp = new int[nums.Length];
+        dp[0] = nums[0];
+        dp[1] = Math.Max(nums[0], nums[1]);
+
+        for (var i = 2; i < dp.Length; i++) {
+            dp[i] = Math.Max(dp[i - 1], dp[i - 2] + nums[i]);
+        }
+
+        return dp.Last();
+    }
+}
+```
+
+## JavaScript
+```javascript
+var rob = function(nums) {
+  if (nums.length === 1) {
+    return nums[0]
+  }
+
+  const dp = Array(nums.length).fill(0)
+  dp[0] = nums[0]
+  dp[1] = Math.max(nums[0], nums[1])
+
+  for (let i = 2; i < dp.length; i++) {
+    dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i])
+  }
+
+  return dp.at(-1)
+};
+```
+
+## Go
+```go
+func rob(nums []int) int {
+    if len(nums) == 1 {
+        return nums[0]
+    }
+
+    dp := make([]int, len(nums))
+    dp[0] = nums[0]
+    dp[1] = max(nums[0], nums[1])
+
+    for i := 2; i < len(dp); i++ {
+        dp[i] = max(dp[i-1], dp[i-2]+nums[i])
+    }
+
+    return dp[len(dp)-1]
+}
+```
+
+## Ruby
+```ruby
+def rob(nums)
+  return nums[0] if nums.size == 1
+
+  dp = Array.new(nums.size, 0)
+  dp[0] = nums[0]
+  dp[1] = [ nums[0], nums[1] ].max
+
+  (2...dp.size).each do |i|
+    dp[i] = [ dp[i - 1], dp[i - 2] + nums[i] ].max
+  end
+
+  dp[-1]
+end
+```
