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feat: update leetcode solutions: No.0198, No.0213
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‎lcof/README.md

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# [《剑指 Offer（第 2 版）》系列](https://leetcode-cn.com/problemset/lcof/)
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![](./lcof.png)
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本书精选谷歌、微软等知名 IT 企业的典型面试题，系统地总结了如何在面试时写出高质量代码，如何优化代码效率，以及分析、解决难题的常用方法。
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[English Version](/lcof/README_EN.md)

‎lcof/README_TEMPLATE.md

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# [题目](这里是题目链接，如：https://leetcode-cn.com/problems/shu-zu-zhong-zhong-fu-de-shu-zi-lcof/)
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# [{}. {}]({})
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[English Version]({})
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## 题目描述
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<!-- 这里写题目描述 -->
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{}
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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### Python3
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### Java
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### ...
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### **...**
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```
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```
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<!-- tabs:end -->

‎lcof/lcof.png

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‎lcof/lcof_list.json

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‎solution/0100-0199/0198.House Robber/README.md

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@@ -28,6 +28,8 @@
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<!-- 这里可写通用的实现逻辑 -->
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动态规划法。状态转移方程：`f(n) = Math.max(f(n - 2) + nums[n], nums[n - 1])`。
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def rob(self, nums: List[int]) -> int:
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def robRange(nums, start, end):
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if end - start == 0:
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return nums[start]
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pre, cur = 0, nums[start]
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for i in range(start + 1, end + 1):
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pre, cur = cur, max(pre + nums[i], cur)
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return cur
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if not nums:
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return 0
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n = len(nums)
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if n == 1:
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return nums[0]
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pre, cur = nums[0], max(nums[0], nums[1])
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for i in range(2, n):
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t = max(pre + nums[i], cur)
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pre, cur = cur, t
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return cur
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return robRange(nums, 0, len(nums) - 1)
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```
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### **Java**
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class Solution {
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public int rob(int[] nums) {
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int n;
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if (nums == null || (n = nums.length) == 0) {
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return 0;
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}
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if (n == 1) {
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return nums[0];
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}
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int pre = nums[0];
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int cur = Math.max(nums[0], nums[1]);
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for (int i = 2; i < n; ++i) {
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if ((n = nums.length) == 0) return 0;
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return robRange(nums, 0, n - 1);
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}
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private int robRange(int[] nums, int start, int end) {
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if (end - start == 0) return nums[start];
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int pre = 0;
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int cur = nums[start];
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for (int i = start + 1; i < end + 1; ++i) {
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int t = Math.max(pre + nums[i], cur);
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pre = cur;
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cur = t;
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int rob(vector<int>& nums) {
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int n;
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if ((n = nums.size()) == 0) return 0;
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return robRange(nums, 0, n - 1);
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}
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private:
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int robRange(vector<int>& nums, int start, int end) {
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if (end - start == 0) return nums[start];
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int pre = 0;
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int cur = nums[start];
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for (int i = start + 1; i < end + 1; ++i) {
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int t = max(pre + nums[i], cur);
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pre = cur;
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cur = t;
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}
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return cur;
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}
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};
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```
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### **Go**
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```go
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func rob(nums []int) int {
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n := len(nums)
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if n == 0 {
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return 0
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}
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return robRange(nums, 0, n - 1)
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}
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func robRange(nums[]int, start int, end int) int {
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if end - start == 0 {
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return nums[start]
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}
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pre, cur := 0, nums[start]
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for i := start + 1; i < end + 1; i++ {
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pre, cur = cur, max(pre + nums[i], cur)
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}
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return cur
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}
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func max(a, b int) int {
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if (a > b) {
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return a
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}
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return b
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}
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```
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### **...**
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```

‎solution/0100-0199/0198.House Robber/README_EN.md

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@@ -43,16 +43,16 @@
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```python
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class Solution:
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def rob(self, nums: List[int]) -> int:
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def robRange(nums, start, end):
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if end - start == 0:
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return nums[start]
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pre, cur = 0, nums[start]
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for i in range(start + 1, end + 1):
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pre, cur = cur, max(pre + nums[i], cur)
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return cur
4653
if not nums:
4754
return 0
48-
n = len(nums)
49-
if n == 1:
50-
return nums[0]
51-
pre, cur = nums[0], max(nums[0], nums[1])
52-
for i in range(2, n):
53-
t = max(pre + nums[i], cur)
54-
pre, cur = cur, t
55-
return cur
55+
return robRange(nums, 0, len(nums) - 1)
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```
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### **Java**
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class Solution {
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public int rob(int[] nums) {
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int n;
64-
if (nums == null || (n = nums.length) == 0) {
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return 0;
66-
}
67-
if (n == 1) {
68-
return nums[0];
69-
}
70-
int pre = nums[0];
71-
int cur = Math.max(nums[0], nums[1]);
72-
for (int i = 2; i < n; ++i) {
64+
if ((n = nums.length) == 0) return 0;
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return robRange(nums, 0, n - 1);
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}
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private int robRange(int[] nums, int start, int end) {
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if (end - start == 0) return nums[start];
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int pre = 0;
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int cur = nums[start];
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for (int i = start + 1; i < end + 1; ++i) {
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int t = Math.max(pre + nums[i], cur);
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pre = cur;
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cur = t;
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}
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```
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### **C++**
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```cpp
85+
class Solution {
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public:
87+
int rob(vector<int>& nums) {
88+
int n;
89+
if ((n = nums.size()) == 0) return 0;
90+
return robRange(nums, 0, n - 1);
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}
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private:
94+
int robRange(vector<int>& nums, int start, int end) {
95+
if (end - start == 0) return nums[start];
96+
int pre = 0;
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int cur = nums[start];
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for (int i = start + 1; i < end + 1; ++i) {
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int t = max(pre + nums[i], cur);
100+
pre = cur;
101+
cur = t;
102+
}
103+
return cur;
104+
}
105+
};
106+
```
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### **Go**
109+
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```go
111+
func rob(nums []int) int {
112+
n := len(nums)
113+
if n == 0 {
114+
return 0
115+
}
116+
return robRange(nums, 0, n - 1)
117+
}
118+
119+
func robRange(nums[]int, start int, end int) int {
120+
if end - start == 0 {
121+
return nums[start]
122+
}
123+
pre, cur := 0, nums[start]
124+
for i := start + 1; i < end + 1; i++ {
125+
pre, cur = cur, max(pre + nums[i], cur)
126+
}
127+
return cur
128+
}
129+
130+
func max(a, b int) int {
131+
if (a > b) {
132+
return a
133+
}
134+
return b
135+
}
136+
```
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### **...**
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```

‎solution/0100-0199/0198.House Robber/Solution.cpp

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class Solution {
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public:
33
int rob(vector<int>& nums) {
4-
if (nums.size() == 0)
5-
return 0 ;
6-
if (nums.size() == 1)
7-
return nums[0] ;
8-
vector<int> dp(nums.size(), -1) ;
9-
dp[0] = nums[0] ;
10-
dp[1] = max(nums[0], nums[1]) ;
11-
for (int i = 2; i < nums.size(); ++i)
12-
{
13-
dp[i] = max(
14-
dp[i-2] + nums[i],
15-
dp[i-1]
16-
) ;
4+
int n;
5+
if ((n = nums.size()) == 0) return 0;
6+
return robRange(nums, 0, n - 1);
7+
}
8+
9+
private:
10+
int robRange(vector<int>& nums, int start, int end) {
11+
if (end - start == 0) return nums[start];
12+
int pre = 0;
13+
int cur = nums[start];
14+
for (int i = start + 1; i < end + 1; ++i) {
15+
int t = max(pre + nums[i], cur);
16+
pre = cur;
17+
cur = t;
1718
}
18-
return dp[dp.size()-1] ;
19+
return cur;
1920
}
2021
};

‎solution/0100-0199/0198.House Robber/Solution.go

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1-
/*
2-
198.House Robber
3-
动态规划的思路，找到状态转移方程 `f(n) = max(f(n-2)+nums[n], f(n-1))` ，初始值为 `f(0)=0, f(1)=nums[1]`
4-
*/
5-
61
func rob(nums []int) int {
7-
x, y := 0, 0
8-
for _, n := range nums {
9-
x, y = y, x+n
10-
if x > y {
11-
y = x
12-
}
13-
}
14-
return y
2+
n := len(nums)
3+
if n == 0 {
4+
return 0
5+
}
6+
return robRange(nums, 0, n - 1)
7+
}
8+
9+
func robRange(nums[]int, start int, end int) int {
10+
if end - start == 0 {
11+
return nums[start]
12+
}
13+
pre, cur := 0, nums[start]
14+
for i := start + 1; i < end + 1; i++ {
15+
pre, cur = cur, max(pre + nums[i], cur)
16+
}
17+
return cur
18+
}
19+
20+
func max(a, b int) int {
21+
if (a > b) {
22+
return a
23+
}
24+
return b
1525
}

‎solution/0100-0199/0198.House Robber/Solution.java

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Original file line numberDiff line numberDiff line change
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class Solution {
22
public int rob(int[] nums) {
33
int n;
4-
if (nums == null || (n = nums.length) == 0) {
5-
return 0;
6-
}
7-
if (n == 1) {
8-
return nums[0];
9-
}
10-
int pre = nums[0];
11-
int cur = Math.max(nums[0], nums[1]);
12-
for (int i = 2; i < n; ++i) {
4+
if ((n = nums.length) == 0) return 0;
5+
return robRange(nums, 0, n - 1);
6+
}
7+
8+
private int robRange(int[] nums, int start, int end) {
9+
if (end - start == 0) return nums[start];
10+
int pre = 0;
11+
int cur = nums[start];
12+
for (int i = start + 1; i < end + 1; ++i) {
1313
int t = Math.max(pre + nums[i], cur);
1414
pre = cur;
1515
cur = t;

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