Sicily 1543. Completing Brackets 解题报告

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

A series of brackets is complete if we can pair off each left bracket '[' with a right bracket ']' that occurs later in the series. Every bracket must participate in exactly one such pair.

Given a String text add the minimal number of brackets to the beginning and/or end of text to make it complete. Return the result.

Input

Each line of input contains a string, text. text will have between 1 and 50 characters inclusive, and contain only the characters "[" and "]". Process to the end of input.

Output

Output one line for each case, the result.

Sample Input

[[
][
[[[[]]]]

Sample Output

[[]]
[][]
[[[[]]]]

 思路：

其实是非常简单的一道题，一开始没注意看题目要求只能在开头或者结尾加上括号把问题复杂化了。

由于这里已经给定了是'['和']'两个符号，所以连stack都不需要，直接用一个count来计数，省事！

这里count初始化为0，当读到一个左括号时，计数器加1，读到右括号时，计时器减1，而且只要count小于0则马上加1并输出一个'[' （在左边补上数量不够的左括号），读取完后再输出count个的右括号即可。

#include<iostream>
using namespace std;

int main(){
    string s;
    while(cin>>s){
        int count=0;
        int len=s.length();
        for(int i=0;i<len;i++){
            if(s[i]=='[')
                count++;
            else{
                count--;
            }
            if(count<0){
                cout<<'[';
                count++;
            }
        }
        cout<<s;
        while(count>0){
            cout<<']';
            count--;
        }
        cout<<endl;
    }
    return 0;
}