判断两个链表有无交点，如果有请给出交点
/* * if list_a and list_b has cross point return the addrss of cross-point. * else return NULL */static List *has_cross(List *list_a, List *list_b){ List *pa; List *pb; int len_a, len_b; int i; len_a = len_b = 0; pa = list_a; /* 遍历链表a，并记录下此链表的长度 */ while (pa->next != NULL) { pa = pa->next; len_a++; } pb = list_b; /* 遍历链表b，并记录下链表b的长度 */ while (pb->next != NULL) { pb = pb->next; len_b++; } if (pa != pb) /* 如果指针pa，pb最后不相等，则两链表没有焦点 */ return NULL; /* 较长的链表先向后调整到第N个元素(N = max(len_a, len_b) - min(len_a, len_b))然后两链表同步向后调整，同时比较地址是否相等，如果相等则为首焦点。*/ pa = list_a; pb = list_b; if (len_a > len_b) { for (i = len_a - len_b; i; i--) pa = pa->next; } else { for (i = len_b - len_a; i; i--) pb = pb->next; } while (pa != pb) { pa = pa->next; pb = pb->next; } return pa;}
页:
[1]