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[pull] master from labuladong:master #100

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Merged
pull merged 3 commits into AlgorithmAndLeetCode:master from labuladong:master
Jun 13, 2023
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101 changes: 76 additions & 25 deletions 多语言解法代码/solution_code.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -2624,42 +2624,41 @@ func backtrack(left, right int, track *string, res *[]string) {
```

```java
// by labuladong (java)
// by chatGPT (java)
class Solution {
public:
vector<string> generateParenthesis(int n) {
if (n == 0) return {};
public List<String> generateParenthesis(int n) {
if (n == 0) return new ArrayList<>();
// 记录所有合法的括号组合
vector<string> res;
List<String> res = new ArrayList<>();
// 回溯过程中的路径
string track;
StringBuilder track = new StringBuilder();
// 可用的左括号和右括号数量初始化为 n
backtrack(n, n, track, res);
return res;
}

// 可用的左括号数量为 left 个,可用的右括号数量为 rgiht 个
// 可用的左括号数量为 left 个,可用的右括号数量为 right 个
void backtrack(int left, int right,
string& track, vector<string>& res) {
StringBuilder track, List<String> res) {
// 若左括号剩下的多,说明不合法
if (right < left) return;
// 数量小于 0 肯定是不合法的
if (left < 0 || right < 0) return;
// 当所有括号都恰好用完时,得到一个合法的括号组合
if (left == 0 && right == 0) {
res.push_back(track);
res.add(track.toString());
return;
}

// 尝试放一个左括号
track.push_back('('); // 选择
track.append('('); // 选择
backtrack(left - 1, right, track, res);
track.pop_back(); // 撤消选择
track.deleteCharAt(track.length() - 1); // 撤消选择

// 尝试放一个右括号
track.push_back(')'); // 选择
track.append(')'); // 选择
backtrack(left, right - 1, track, res);
track.pop_back(); // 撤消选择
track.deleteCharAt(track.length() - 1); // 撤消选择
}
}
```
Expand Down Expand Up @@ -26417,8 +26416,9 @@ class Solution:
for v in nums:
valToFreq[v] = valToFreq.get(v, 0) + 1

# 优先队列按照键值对中的值(元素出现频率)从小到大排序
pq = [(freq, val) for val, freq in valToFreq.items()]
# 二叉堆按照键值对中的值(元素出现频率的负值)从小到大排列
# 从二叉堆中pop出来的就是频率最大的键 (频率越大,负值越小)
pq = [(-freq, val) for val, freq in valToFreq.items()]
heapq.heapify(pq)

# 将前 k 个最大元素装入 res
Expand Down Expand Up @@ -26872,7 +26872,7 @@ func backtrack(left int, right int, track *string, res *[]string) {
```

```java
// by labuladong (java)
// by chatGPT (java)
class Solution {
public List<String> generateParenthesis(int n) {
if (n == 0) return new ArrayList<>();
Expand Down Expand Up @@ -32964,6 +32964,53 @@ class Solution:

https://leetcode.cn/problems/lMSNwu 的多语言解法👆

https://leetcode.cn/problems/letter-case-permutation 的多语言解法👇

```java
// by labuladong (java)
class Solution {
public List<String> letterCasePermutation(String s) {
backtrack(s, 0);
return res;
}

StringBuilder track = new StringBuilder();
List<String> res = new LinkedList<>();

void backtrack(String s, int index) {
if (index == s.length()) {
res.add(track.toString());
return;
}

if ('0' <= s.charAt(index) && s.charAt(index) <= '9') {
// s[index] 是数字
// 做选择
track.append(s.charAt(index));
backtrack(s, index + 1);
// 撤销选择
track.deleteCharAt(track.length() - 1);
} else {
// s[index] 是字母

// 小写字母,做选择
track.append(Character.toLowerCase(s.charAt(index)));
backtrack(s, index + 1);
// 撤销选择
track.deleteCharAt(track.length() - 1);

// 大写字母,做选择
track.append(Character.toUpperCase(s.charAt(index)));
backtrack(s, index + 1);
// 撤销选择
track.deleteCharAt(track.length() - 1);
}
}
}
```

https://leetcode.cn/problems/letter-case-permutation 的多语言解法👆

https://leetcode.cn/problems/letter-combinations-of-a-phone-number 的多语言解法👇

```cpp
Expand Down Expand Up @@ -68764,39 +68811,43 @@ MyQueue.prototype.empty = function() {

```python
# by chatGPT (python)
class MyQueue:
class CQueue:
def __init__(self):
"""
初始化一个队列,使用两个堆栈 s1 和 s2
"""
self.s1 = []
self.s2 = []

def push(self, x: int) -> None:
def appendTail(self, value):
"""
添加元素到队尾
"""
self.s1.append(x)
self.s1.append(value)

def pop(self) -> int:
def deleteHead(self):
"""
删除队头的元素并返回
"""
# 先调用 peek 保证 s2 非空
self.peek()
return self.s2.pop()
if self.peek() is None:
return -1
else:
return self.s2.pop()

def peek(self) -> int:
def peek(self):
"""
返回队头元素
"""
if not self.s2:
# 把 s1 元素压入 s2
while self.s1:
self.s2.append(self.s1.pop())
return self.s2[-1]

return self.s2[-1] if self.s2 else None


def empty(self) -> bool:
def empty(self):
"""
判断队列是否为空
"""
Expand Down
1 change: 1 addition & 0 deletions 算法思维系列/回溯算法详解修订版.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -445,6 +445,7 @@ def backtrack(...):
| [698. Partition to K Equal Sum Subsets](https://leetcode.com/problems/partition-to-k-equal-sum-subsets/?show=1) | [698. 划分为k个相等的子集](https://leetcode.cn/problems/partition-to-k-equal-sum-subsets/?show=1) |
| [77. Combinations](https://leetcode.com/problems/combinations/?show=1) | [77. 组合](https://leetcode.cn/problems/combinations/?show=1) |
| [78. Subsets](https://leetcode.com/problems/subsets/?show=1) | [78. 子集](https://leetcode.cn/problems/subsets/?show=1) |
| [784. Letter Case Permutation](https://leetcode.com/problems/letter-case-permutation/?show=1) | [784. 字母大小写全排列](https://leetcode.cn/problems/letter-case-permutation/?show=1) |
| [93. Restore IP Addresses](https://leetcode.com/problems/restore-ip-addresses/?show=1) | [93. 复原 IP 地址](https://leetcode.cn/problems/restore-ip-addresses/?show=1) |
| - | [剑指 Offer 34. 二叉树中和为某一值的路径](https://leetcode.cn/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof/?show=1) |
| - | [剑指 Offer II 079. 所有子集](https://leetcode.cn/problems/TVdhkn/?show=1) |
Expand Down

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