链表专题

1、uva 11988 Broken Keyboard

　　题意：给出键盘打字的顺序，遇到[到句首，遇到]倒句末。求最后输出的句子。

　　思路：

①单向链表数组模拟

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 const int maxn = 100100;
 6 char s[maxn];
 7 int Next[maxn];
 8 int main()
 9 {
10     while (~scanf("%s", s + 1))
11     {
12         int len = strlen(s + 1);
13         Next[0] = 0;
14         int cur = 0, last = 0;
15         for (int i = 1; i <= len; i++)
16         {
17             if (s[i] == '[')cur=0 ;
18             else if (s[i] == ']') cur = last;
19             else
20             {
21                 Next[i] = Next[cur];
22                 Next[cur] = i;
23                 if (last == cur) last = i;
24                 cur = i;
25             }
26         }
27         for (int i = Next[0]; i != 0; i = Next[i]) printf("%c", s[i]);
28         printf("\n");
29     }
30     return 0;
31 }
View Code

②双向链表数组模拟

 1 #include<iostream>
 2 #include<cstring>
 3 const int maxn = 100100;
 4 char s[maxn];
 5 int L[maxn], R[maxn];
 6 int main()
 7 {
 8     while (~scanf("%s", s+1))
 9     {
10         int len = strlen(s+1);
11         L[0] = R[0] = 0;
12         for (int i = 1; i <= len; i++)
13         {
14             if (s[i] == '[') L[i] = 0, R[i] = R[0];
15             else if (s[i] == ']') L[i] = L[0], R[i] = 0;
16             else
17             {
18                 L[i] = i - 1, R[i] = R[i - 1];
19             }
20             L[R[i]] = i, R[L[i]] = i;
21         }
22         for (int i = R[0]; i != 0; i = R[i])
23         {
24             if (s[i] == '[' || s[i] == ']')continue;
25             printf("%c", s[i]);
26         }
27         printf("\n");
28     }
29     return 0;
30 }
View Code

③STL list使用

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<list>
 5 using namespace std;
 6 const int maxn = 100100;
 7 char s[maxn];
 8 
 9 int main()
10 {
11     while (~scanf("%s", s))
12     {
13         int len = strlen(s);
14         list<char>lt;
15         list<char>::iterator pos=lt.end();
16         for (int i = 0; i < len; i++)
17         {
18             if (s[i] == '[') pos=lt.begin();
19             else if (s[i] == ']') pos=lt.end();
20             else
21             {
22                 lt.insert(pos, s[i]);
23             }
24         }
25         for (pos = lt.begin(); pos != lt.end(); pos++)
26         {
27             printf("%c", *pos);
28         }
29         printf("\n");
30     }
31     return 0;
32 }
View Code

 

posted @ 2017-08-25 22:30  萌萌的美男子  阅读(148)  评论(0)  编辑  收藏  举报