目录
- 1 介绍
- 2 训练
- 3 参考
1 介绍
本博客用来记录代码随想录leetcode200题中栈与队列部分的题目。
2 训练
题目1：232. 用栈实现队列
C++代码如下，
#include <stack>
class MyQueue {
private:
stack<int> a;
stack<int> b; //辅助栈
public:
MyQueue() {
a = stack<int>();
b = stack<int>();
}
void push(int x) {
//先把a中的内容转移到b
while (!a.empty()) {
b.push(a.top());
a.pop();
}
//再将x推入a中
a.push(x);
//最后将b中的内容转移回a
while (!b.empty()) {
a.push(b.top());
b.pop();
}
return;
}
int pop() {
int t = a.top();
a.pop();
return t;
}
int peek() {
return a.top();
}
bool empty() {
return a.empty();
}
};
/**
* Your MyQueue object will be instantiated and called as such:
* MyQueue* obj = new MyQueue();
* obj->push(x);
* int param_2 = obj->pop();
* int param_3 = obj->peek();
* bool param_4 = obj->empty();
*/
python3代码如下，
class MyQueue:
def __init__(self):
self.a = []
self.b = []
def push(self, x: int) -> None:
#将a中内容转移到b
while len(self.a) > 0:
self.b.append(self.a[-1])
self.a.pop()
#将x推入到a中
self.a.append(x)
#将b中内容转移回a
while len(self.b) > 0:
self.a.append(self.b[-1])
self.b.pop()
def pop(self) -> int:
return self.a.pop()
def peek(self) -> int:
return self.a[-1]
def empty(self) -> bool:
return len(self.a) == 0
# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()
题目2：225. 用队列实现栈
C++代码如下，
class MyStack {
private:
queue<int> a;
queue<int> b; //辅助队列
public:
MyStack() {
a = queue<int>();
b = queue<int>();
}
void push(int x) {
//将a中内容转移到b
while (!a.empty()) {
b.push(a.front());
a.pop();
}
//将x推入a中
a.push(x);
//将b中内容转移回a
while (!b.empty()) {
a.push(b.front());
b.pop();
}
return;
}
int pop() {
int t = a.front();
a.pop();
return t;
}
int top() {
return a.front();
}
bool empty() {
return a.empty();
}
};
/**
* Your MyStack object will be instantiated and called as such:
* MyStack* obj = new MyStack();
* obj->push(x);
* int param_2 = obj->pop();
* int param_3 = obj->top();
* bool param_4 = obj->empty();
*/
python3代码如下，
import queue
class MyStack:
def __init__(self):
self.a = queue.Queue()
self.b = queue.Queue() #辅助队列
def push(self, x: int) -> None:
#将a中的内容转移到b
while not self.a.empty():
self.b.put(self.a.get())
#往a中推入x
self.a.put(x)
#将b中的内容转移回a
while not self.b.empty():
self.a.put(self.b.get())
return
def pop(self) -> int:
return self.a.get()
def top(self) -> int:
t = self.a.get()
self.push(t)
return t
def empty(self) -> bool:
return self.a.empty()
# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()
题目3：20. 有效的括号
C++代码如下，
class Solution {
public:
bool isValid(string s) {
stack<char> stk;
for (auto c : s) {
if (c == '(' || c == '[' || c == '{') {
stk.push(c);
} else {
if (!stk.empty()) {
char t = stk.top();
stk.pop();
if ((t == '(' && c == ')')
|| (t == '[' && c == ']')
|| (t == '{' && c == '}')) {
//
} else {
return false;
}
} else {
return false;
}
}
}
return stk.empty();
}
};
python3代码如下，
class Solution:
def isValid(self, s: str) -> bool:
stk = []
for c in s:
if c in "([{":
stk.append(c)
else:
if len(stk) == 0:
return False
else:
t = stk.pop()
if (t == '(' and c == ')') or \
(t == '[' and c == ']') or \
(t == '{' and c == '}'):
pass
else:
return False
return len(stk) == 0
题目4：1047. 删除字符串中的所有相邻重复项
C++代码如下，
class Solution {
public:
string removeDuplicates(string s) {
stack<char> stk;
for (auto c : s) {
if (stk.empty() || stk.top() != c) {
stk.push(c);
} else {
stk.pop();
}
}
string t;
while (!stk.empty()) {
t += stk.top();
stk.pop();
}
reverse(t.begin(), t.end());
return t;
}
};
python3代码如下，
class Solution:
def removeDuplicates(self, s: str) -> str:
stk = []
for c in s:
if len(stk) == 0 or stk[-1] != c:
stk.append(c)
else:
stk.pop()
res = "".join(stk)
return res
题目5：150. 逆波兰表达式求值
C++代码如下，
class Solution {
public:
int evalRPN(vector<string>& tokens) {
stack<int> stk;
string operations = "+-*/";
for (auto token : tokens) {
if (operations.find(token) == string::npos) {
stk.push(stoi(token));
} else {
int b = stk.top();
stk.pop();
int a = stk.top();
stk.pop();
if (token == "+") stk.push(a+b);
else if (token == "-") stk.push(a-b);
else if (token == "*") stk.push(a*b);
else stk.push(a/b);
}
}
return stk.top();
}
};
python3代码如下，
class Solution:
def evalRPN(self, tokens: List[str]) -> int:
stk = []
for token in tokens:
if token not in ["+", "-", "*", "/"]:
stk.append(int(token))
else:
b = stk[-1]
stk.pop()
a = stk[-1]
stk.pop()
if token == "+":
stk.append(a+b)
elif token == "-":
stk.append(a-b)
elif token == "*":
stk.append(int(a*b))
else:
stk.append(int(a/b)) #注意a//b与int(a/b)的区别
return stk[-1]
题目6：239. 滑动窗口最大值
C++代码如下，
python3代码如下，
3 参考
代码随想录官网
