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做网站设计需要什么软件,如何找网站推广,郑州哪个公司专业做网站,门户网站开发需要(100分)- 表达式括号匹配#xff08;Java JS Python C#xff09;题目描述(1(23)*(3(80))1-2)这是一个简单的数学表达式,今天不是计算它的值,而是比较它的括号匹配是否正确。前面这个式子可以简化为(()(()))这样的括号我们认为它是匹配正确的,而((())这样的…(100分)- 表达式括号匹配Java JS Python C题目描述(1(23)*(3(80))1-2)这是一个简单的数学表达式,今天不是计算它的值,而是比较它的括号匹配是否正确。前面这个式子可以简化为(()(()))这样的括号我们认为它是匹配正确的,而((())这样的我们就说他是错误的。注意括号里面的表达式可能是错的,也可能有多个空格对于这些我们是不用去管的我们只关心括号是否使用正确。输入描述给出一行表达式(长度不超过 100)。输出描述如果匹配正确输出括号的对数否则输出-1。用例输入(1(23)*(3(80))1-2)输出4说明无题目解析本题就是括号匹配的变种题只是加入了一些干扰字符我们可以用正则去掉非()的字符然后利用栈结构校验括号是否成对JavaScript算法源码/* JavaScript Node ACM模式 控制台输入获取 */ const readline require(readline); const rl readline.createInterface({ input: process.stdin, output: process.stdout, }); rl.on(line, (line) { let count 0; const stack []; for (let c of line) { if (c ! ( c ! )) continue; if (stack.length c )) { if (stack.at(-1) () { stack.pop(); count; continue; } else { return console.log(-1); } } stack.push(c); } if (stack.length) return console.log(-1); return console.log(count); });Java算法源码import java.util.LinkedList; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc new Scanner(System.in); String s sc.nextLine(); System.out.println(getResult(s)); } public static int getResult(String s) { int count 0; LinkedListCharacter stack new LinkedList(); for (int i 0; i s.length(); i) { char c s.charAt(i); if (c ! ) c ! () continue; if (stack.size() 0 c )) { if (stack.getLast() () { stack.removeLast(); count; continue; } return -1; } stack.add(c); } if (stack.size() 0) return -1; return count; } }Python算法源码# 输入获取 s input() # 算法入口 def getResult(s): count 0 stack [] for c in s: if c ! ( and c ! ): continue if len(stack) 0 and c ): if stack[-1] (: stack.pop() count 1 continue return -1 stack.append(c) if len(stack) 0: return -1 return count # 算法调用 print(getResult(s))C算法源码#include stdio.h #include stdlib.h #include string.h typedef char E; typedef struct ListNode { E ele; struct ListNode *prev; struct ListNode *next; } ListNode; typedef struct { int size; ListNode *head; ListNode *tail; } LinkedList; LinkedList *new_LinkedList(); void addLast_LinkedList(LinkedList *link, E ele); E removeLast_LinkedList(LinkedList *link); E getLast_LinkedList(LinkedList* link); int main() { char s[100]; gets(s); int count 0; LinkedList *stack new_LinkedList(); for (int i 0; i strlen(s); i) { if(s[i] ! ) s[i] ! () continue; if(stack-size 0 s[i] )) { if(getLast_LinkedList(stack) () { removeLast_LinkedList(stack); count; continue; } puts(-1); return 0; } addLast_LinkedList(stack,s[i]); } if(stack-size 0) { puts(-1); } else { printf(%d\n, count); } return 0; } LinkedList *new_LinkedList() { LinkedList *link (LinkedList *) malloc(sizeof(LinkedList)); link-size 0; link-head NULL; link-tail NULL; return link; } void addLast_LinkedList(LinkedList *link, E ele) { ListNode *node (ListNode *) malloc(sizeof(ListNode)); node-ele ele; node-prev NULL; node-next NULL; if (link-size 0) { link-head node; link-tail node; } else { link-tail-next node; node-prev link-tail; link-tail node; } link-size; } E removeLast_LinkedList(LinkedList *link) { if (link-size 0) exit(-1); ListNode *removed link-tail; if (link-size 1) { link-head NULL; link-tail NULL; } else { link-tail link-tail-prev; link-tail-next NULL; } link-size--; E res removed-ele; free(removed); return res; } E getLast_LinkedList(LinkedList* link) { if(link-size 0) exit(-1); return link-tail-ele; }