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阅读以下说明,回答问题1~5,将解答填入对应的解答栏内。
[说明] 若s和t是用单链表存储的两个串,设计一个函数将s串中首次与串t匹配的字串逆置。
linkstring * invert - substring ( s, t)
linkstring * s, * t;
{
linkstring *prior, *p, *t1, *r, *q, *u;
prior =s;
p=s;
t1 =t;
if ( (1) ) printf ( "error\n") ;
else
{
while { p ! = NULL && t1! = NULL)
{
if ( p- >data = = t1 - >data)
{
p = p- >link;
t1 = t1- >link;
}
else
{
(2)
p = prior - > link;
}
t1 = t- >link;
}
if ( t1 ! : NULL) printf ("cannot find");
else
{
(3)
r = q- >link;
q- >link = p;
while (r ! = p)
{
u = r- >link;
(4)
q=r;
r = u;
}
(5)
}
}
}
参考答案
更多 “ 阅读以下说明,回答问题1~5,将解答填入对应的解答栏内。[说明] 若s和t是用单链表存储的两个串,设计一个函数将s串中首次与串t匹配的字串逆置。linkstring * invert - substring ( s, t)linkstring * s, * t;{linkstring *prior, *p, *t1, *r, *q, *u;prior =s;p=s;t1 =t;if ( (1) ) printf ( "error\n") ;else{while { p ! = NULL t1! = NULL){if ( p- >data = = t1 - >data){p = p- >link;t1 = t1- >link;}else{(2)p = prior - > link;}t1 = t- >link;}if ( t1 ! : NULL) printf ("cannot find");else{(3)r = q- >link;q- >link = p;while (r ! = p){u = r- >link;(4)q=r;r = u;}(5)}}} ” 相关考题
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