Output

For each case, print the Beiju text on the screen.

思路：使用連結串列來模擬，遇到Home鍵，就將後邊的文字插入到這段文字的最前邊，遇到End鍵，就插入到這段文字的最後邊。但是用連結串列會用到指標，過程比較繁瑣。這裡用一個Next陣列模擬指向，Next[i]表示當前顯示屏中s[i]右邊的字元下標。再用一個cur表示當前游標的位置，last表示最後一個字元的記錄位置，這樣遇到End鍵，就能直接找到游標下一個指向的字元位置了。

#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 100005;
char s[maxn];
int next[maxn];
int cur,last;

int main(int argc, char const *argv[])
{
	while(scanf("%s",s+1) != EOF)
	{
		int len = strlen(s+1);
		cur = last = 0;
		next[0] = 0;

		for(int i = 1;i <= len;i++)
		{
			char ch = s[i];
			if(ch == '[')
				cur = 0;
			else if(ch == ']')
				cur = last;
			else
			{
				next[i] = next[cur];
				next[cur] = i;
				if(cur == last)
					last = i;
				cur = i;
			}
		}

		for(int i = next[0];i != 0;i = next[i])
			printf("%c",s[i]);
		printf("\n");
	}
	return 0;
}
 

#include <cstdio>
#include <queue>
using namespace std;

const int maxn = 100010;
char str[maxn];

int main(int argc, char const *argv[])
{
	while(~scanf("%s",str))
	{
		deque<int> q;
		int i = 0;
		while(str[i] == '[' || str[i] == ']')
			i++;
		q.push_front(i);
		while(str[i] != '\0')
		{
			if(str[i] == '[')
			{
				q.push_front(i+1);
				str[i] = '\0';
			}
			else if(str[i] == ']')
			{
				q.push_back(i+1);
				str[i] = '\0';
			}
			i++;
		}

		while(!q.empty())
		{
			printf("%s",str+q.front());
			q.pop_front();
		}
		printf("\n");
	}
	return 0;
}