Out[23]: '瓜南小个一是你' 参考stackflow上的答案。 原理是:This is extended slice syntax. It works by doing [begin:end:step] - by leaving begin and end off and specifying a step of -1, it reverses a string. 切片介绍:切片操作符中的第一个数(
1. Python中栈的表达: stack = [] 2. Python中加入元素到栈,和从栈中弹出元素的方法: #加入栈 stack.append(char) #弹出栈 s[i] = stack.pop() 解法3:使用range class Solution: def reverseString(self, s: List[str]) -> None: n = len(s) for i in range(n//2): s[i], s[n-i-...
Initializing the JIT engine with a stack: >>> jitter=machine.jitter(loc_db,jit_type='python') >>> jitter.init_stack() Add the shellcode in an arbitrary memory location: >>> run_addr=0x40000000>>>frommiasm.jitter.cstsimportPAGE_READ,PAGE_WRITE>>> jitter.vm.add_memory_page(run_addr,...
def reverse_handle(start, end, numstr): stack_ = list() for i in range(start, end): ch = numstr[i] stack_.append(ch) newnumstr = '' for i in range(start, end): newnumstr += stack_.pop() count_0 = 0 for i in range(end-start): if i > 0 and newnumstr[i] == '...
A mobile-first, full-stack, web application for lovers of metal music that aims to link them with other fans of the genre react react-router postgresql-database discogs-api nodemailer-transport multer-storage user-authentication bcrypt-encryption youtube-search-api ticketmaster-api pm2-module express...
Currently tracking 1092326991 bots worldwide Breakpoint 1, 0x0804865c in decrypt () (gdb) n # n 单步步入 Single stepping until exit from function decrypt, which has no line number information. 0x08048725 in authenticate () (gdb) x/5sw $eax # x就是查看内存内容 5表示查看的单元个数 s是...
(&self->data); vec_capacity = self->vec_capacity; retstr->buf = heap_buf->buf; retstr->p_buf_capacity = &heap_buf->buf_capacity; retstr->vec_capacity = vec_capacity; } else // capacity <= 0x100 { stack_buf = smallvec::SmallVecData<A>::inline_mut(&self->data); buf = _...
Or, since we have JTAG access to the chip, we can just put a breakpoint on the correct address in memory and let the MCU do the hard work for us. I will leave the first solution for you to implement as an exercise. It should take more or less 10 lines of Python or C c...
stack<int> cache; for(int i = 0 ; i < tokens.size(); i++){ if(tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/"){ int num2 = cache.top(); cache.pop(); int num1 = cache.top(); cache.pop(); cache.push(calculate...
stack=[]# 列表实现栈的 先入后出功能count=0whilecur:count=count+1stack.append(cur)# 入栈cur=cur.nextifcount%k==0:# 已存储k个元素count=0whilestack:# 出栈倒换tmpresult.next=stack.pop()tmpresult=tmpresult.next# 处理 stack 中剩余元素whilestack:tmpresult.next=stack.pop(0)#tmpresult=tmpres...