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50 | 50 |
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51 | 51 | <!-- 这里可写通用的实现逻辑 -->
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52 | 52 |
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| 53 | +直接利用上一题的代码,同样的思路,只不过需要特殊考虑第一个房子(环的连接处) |
| 54 | + |
53 | 55 | <!-- tabs:start -->
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54 | 56 |
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55 | 57 | ### **Python3**
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56 | 58 |
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57 | 59 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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58 | 60 |
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59 | 61 | ```python
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60 |
| - |
| 62 | +class Solution: |
| 63 | + def rob(self, nums: List[int]) -> int: |
| 64 | + if len(nums) == 1: |
| 65 | + return nums[0] |
| 66 | + return max(self._rob(nums[:-1]), self._rob(nums[1:])) |
| 67 | + |
| 68 | + def _rob(self, nums: List[int]) -> int: |
| 69 | + if len(nums) == 1: |
| 70 | + return nums[0] |
| 71 | + |
| 72 | + n = len(nums) |
| 73 | + dp = [0] * n |
| 74 | + dp[0], dp[1] = nums[0], max(nums[0], nums[1]) |
| 75 | + for i in range(2, n): |
| 76 | + dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]) |
| 77 | + return dp[n - 1] |
61 | 78 | ```
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62 | 79 |
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63 | 80 | ### **Java**
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64 | 81 |
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65 | 82 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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66 | 83 |
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67 | 84 | ```java
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| 85 | +class Solution { |
| 86 | + public int rob(int[] nums) { |
| 87 | + if (nums.length == 1) { |
| 88 | + return nums[0]; |
| 89 | + } |
| 90 | + return Math.max(_rob(nums, 0, nums.length - 1), _rob(nums, 1, nums.length)); |
| 91 | + } |
| 92 | + |
| 93 | + public int _rob(int[] nums, int start, int end) { |
| 94 | + if (start + 1 == end) { |
| 95 | + return nums[start]; |
| 96 | + } |
| 97 | + int n = end - start; |
| 98 | + int[] dp = new int[n]; |
| 99 | + dp[0] = nums[start]; |
| 100 | + dp[1] = Math.max(nums[start], nums[start + 1]); |
| 101 | + for (int i = 2; i < n; i++) { |
| 102 | + dp[i] = Math.max(dp[i - 2] + nums[start + i], dp[i - 1]); |
| 103 | + } |
| 104 | + return dp[n - 1]; |
| 105 | + } |
| 106 | +} |
| 107 | +``` |
| 108 | + |
| 109 | +### **C++** |
| 110 | + |
| 111 | +```cpp |
| 112 | +class Solution { |
| 113 | +public: |
| 114 | + int rob(vector<int>& nums) { |
| 115 | + if (nums.size() == 1) { |
| 116 | + return nums[0]; |
| 117 | + } |
| 118 | + return max(_rob(nums, 0, nums.size() - 1), _rob(nums, 1, nums.size())); |
| 119 | + } |
| 120 | + |
| 121 | + int _rob(vector<int>& nums, int start, int end) { |
| 122 | + if (start + 1 == end) { |
| 123 | + return nums[start]; |
| 124 | + } |
| 125 | + int n = end - start; |
| 126 | + vector<int> dp(n, 0); |
| 127 | + dp[0] = nums[start]; |
| 128 | + dp[1] = max(nums[start], nums[start + 1]); |
| 129 | + for (int i = 2; i < n; ++i) { |
| 130 | + dp[i] = max(dp[i - 2] + nums[start + i], dp[i - 1]); |
| 131 | + } |
| 132 | + return dp[n - 1]; |
| 133 | + } |
| 134 | +}; |
| 135 | +``` |
68 | 136 |
|
| 137 | +### **Go** |
| 138 | +
|
| 139 | +```go |
| 140 | +func rob(nums []int) int { |
| 141 | + if len(nums) == 1 { |
| 142 | + return nums[0] |
| 143 | + } |
| 144 | + return max(_rob(nums[:len(nums)-1]), _rob(nums[1:])) |
| 145 | +} |
| 146 | +
|
| 147 | +func _rob(nums []int) int { |
| 148 | + if len(nums) == 1 { |
| 149 | + return nums[0] |
| 150 | + } |
| 151 | +
|
| 152 | + n := len(nums) |
| 153 | + dp := make([]int, n) |
| 154 | + dp[0] = nums[0] |
| 155 | + dp[1] = max(nums[0], nums[1]) |
| 156 | + for i := 2; i < n; i++ { |
| 157 | + dp[i] = max(dp[i-2]+nums[i], dp[i-1]) |
| 158 | + } |
| 159 | + return dp[n-1] |
| 160 | +} |
| 161 | +
|
| 162 | +func max(x, y int) int { |
| 163 | + if x > y { |
| 164 | + return x |
| 165 | + } |
| 166 | + return y |
| 167 | +} |
69 | 168 | ```
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70 | 169 |
|
71 | 170 | ### **...**
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