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下面是几个常见的链表笔面问题:
[TOC]
很简单:(复杂度O(n))
int list_len(node_t *head){ int i; for (i = 0; head; head = head->next, i++); return i; } 测试:
int main(int argc, const char *argv[]){ node_t d = {"d", 0}, c = {"c", &d}, b = {"b", &c}, a = {"a", &b}; printf("%d\n", list_len(&a));//4 return 0;} 我们多用几个指针就可以在O(n)完成反转任务:
算法:t遍历链表, q记录t的上一个结点, p是一个临时变量用来缓存t的值
void reverse(node_t *head){ node_t *p = 0, *q = 0, *t = 0; for (t = head; t; p = t, t = t->next, p->next = q, q = p); } 测试:
node_t d = { "d", 0}, c = { "c", &d}, b = { "b", &c}, a = { "a", &b}; list_display(&a); reverse(&a); list_display(&d); 算法:2个指针p, q初始化指向头结点.p先跑到k结点处, 然后q再开始跑, 当p跑到最后跑到尾巴时, q正好到达倒数第k个.复杂度O(n)
node_t *_kth(node_t *head, int k){ int i = 0; node_t *p = head, *q = head; for (; p && i < k; p = p->next, i++); if (i < k) return 0; for (; p->next; p = p->next, q = q->next); return q; } 测试:
node_t d = { "d", 0}, c = { "c", &d}, b = { "b", &c}, a = { "a", &b}; printf("_0 :%s _1: %s _2:%s _3:%s\n", _kth(&a, 0)->data, _kth(&a, 1)->data, _kth(&a, 2)->data, _kth(&a, 3)->data); 输出:
_0 :d _1: c _2:b _3:a
找出中间的那个结点
算法:设两个初始化指向头结点的指针p, q.p每次前进两个结点, q每次前进一个结点, 这样当p到达链表尾巴的时候, q到达了中间.复杂度O(n)
node_t *middle(node_t *head){ node_t *p, *q; for (p = q = head; p->next; p = p->next, q = q->next){ p = p->next; if (!(p->next)) break; } return q; } 测试:
node_t e = {"e", 0}, d = {"d", &e}, c = {"c", &d}, b = {"b", &c}, a = {"a", &b}; printf("%s\n", middle(&a)->data); 给你链表的头结点, 逆序打印这个链表.使用递归(即让系统使用栈), 时间复杂度O(n)
void r_display(node_t *t){ if (t){ r_display(t->next); printf("%s", t->data); }} 如果一个链表有环, 那么它肯定只有一个环.(一个相交结点)
算法:设两个指针p, q, 初始化指向头.p以步长2的速度向前跑, q的步长是1.这样, 如果链表不存在环, p和q肯定不会相遇.如果存在环, p和q一定会相遇.(就像两个速度不同的汽车在一个环上跑绝对会相遇).复杂度O(n)
int any_ring(node_t *head){ node_t *p, *q; for (p = q = head;p; p = p->next, q = q->next){ p = p->next; if (!p) break; if (p == q) return 1; //yes } return 0; //fail find} 测试:
node_t e = {"e", 0}, d = {"d", &e}, c = {"c", &d}, b = {"b", &c}, a = {"a", &b}; e.next = &a; printf("%d\n", any_ring(&a));
还是使用俩指针p和q, p扫描的步长为1, q扫描的步长为2.它们的相遇点为图中meet处(在环上).
假设头指针head到入口点entry之间的距离是K.则当q入环的时候, p已经领先了q为: d = K%n(n为环的周长).
我们设meet处相对entry的距离(沿行进方向)为x, 则有
(n-d)+x = 2x (p行进的路程是q的两倍)
解得x = n-d
那么当p和q在meet处相遇的时候, 从head处再发出一个步长为1的指针r, 可以知道, r和q会在entry处相遇!
算法就是:
初始化三个指针p, q, r全部指向head. 然后p以2的速度行进, q以1的速度行进.当p和q相遇的时候, 发出r指针并以1的速度行进, 当p和r相遇返回这个结点.复杂度O(n)
代码:
node_t *find_entry(node_t *head){ node_t *p, *q, *r; for (p = q = head; p; p = p->next, q = q->next){ p = p->next; if (!p) break; if (p == q) break; } if (!p) return 0; //no ring in list for (r = head, q = q->next; q != r; r = r->next, q = q->next); return r; } 测试:
node_t e = { "e", 0}, d = { "d", &e}, c = { "c", &d}, b = { "b", &c}, a = { "a", &b}; e.next = &d; printf("%s\n", find_entry(&a)->data); 算法:两个指针遍历这两个链表,如果他们的尾结点相同,则必定相交.复杂度O(m+n)
代码实现:
int is_intersect(node_t *a, node_t *b){ if (!a || !b) return -1; //a or b is NULL for (; a->next; a = a->next); for (; b->next; b = b->next); return a == b?1:0; //return 1 for yes, 0 for no} 测试代码 :
node_t e = { "e", 0}, d = { "d", &e}, c = { "c", &d}, b = { "b", &c}, a = { "a", &b}; node_t z = { "z", &c}, y = { "y", &z}, x = { "x", &y}; printf("%d\n", is_intersect(&a, &x)); 假设两个链表a,b.a比b长k个结点(k>=0).
那么当a_ptr,b_ptr两个指针同时分别遍历a,b的时候, 必然b_ptr先到达结尾(NULL),而此时a_ptr落后a的尾巴k个结点.
如果此时再从a的头发出一个指针t,继续和a_ptr 一起走,当a_ptr达到结尾(NULL)时,t恰好走了k个结点.此时从b的头发一个指针s, s和t一起走,因为a比b长k个结点,所以,t和s会一起到达交点.
算法便是:
p,q分别遍历链表a,b,假设q先到达NULL,此时从a的头发出一个指针t,当p到达NULL时,从b的头发出s,当s==t的时候即交点.
代码实现: (注,当a,b不相交,函数返回0,即相交在NULL)
node_t *intersect_point(node_t *a, node_t *b){ node_t *p, *q, *k, *t, *s; for (p = a, q = b; p && q; p = p->next, q = q->next); k = (p == 0)?q:p; //k record the pointer not NULL t = (p == 0)?b:a; //if p arrive at tail first, t = b ; else p = a s = (p == 0)?a:b; for (; k; k = k->next, t = t->next); for (; t != s; t = t->next, s = s->next); return t; } 测试
node_t e = { "e", 0}, d = { "d", &e}, c = { "c", &d}, b = { "b", &c}, a = { "a", &b}; node_t z = { "z", &b}, y = { "y", &z}, x = { "x", &y}; printf("%s\n", intersect_point(&a, &x)->data); 其实我很反对这个做法.
不给头结点的时候怎么删除一个结点d:
把d的下一个结点e的数据拷贝到d中,然后删除e
我认为这是个伪删除,并且这个算法无法处理d是最后一个结点的情况
代码实现:
node_t *delete(node_t *d) { node_t *e = d->next; d->data = e->data; d->next = e->next; } 为了打印的整齐性,我们把结点存储的数据类型改为int(我们存放数字)
比如两个链表1->2->3->4->5和6->7->8,我们想要打印这种效果:
1 2 3 4 5 6 7 8
算法:
p和q两个指针分别遍历链表a和b,假如q先到达0(即a比b长),此时由a头发出t,打印完链表a.p继续移动到0,并打印空格.从b头发出指针s打印链表b
代码:
void foo(node_t *a, node_t *b){ node_t *p, *q, *k, *t, *s; for (p = a, q = b; p && q; p = p->next, q = q->next); k = p?p:q; t = p?a:b; s = p?b:a; for (; t; printf("%d ", t->data), t = t->next); printf("\n"); for (; k; printf(" "), k = k->next); for (; s; printf("%d ", s->data), s = s->next); } 测试:
node_t e = { 5, 0}, d = { 4, &e}, c = { 3, &d}, b = { 2, &c}, a = { 1, &b}; node_t o = { 8, 0}, n = { 7, &o}, m = { 6, &n}; foo(&a, &m); 转载地址：http://iqbii.baihongyu.com/