算法中常常会推断两条单链表是否相交，尽管算法简单，但也值得说一下。

代码中有详尽凝视。 Show you the code !

#undef UNICODE#include <iostream>
#include <Windows.h>
#include <cstring>using namespace std;typedef struct _SINGLE_LIST {_SINGLE_LIST *Next;char Tag;
} SINGLE_LIST, *PSINGLE_LIST;void main()
{PSINGLE_LIST Entry = NULL;char TagIndex = 'A';PSINGLE_LIST Head1 = (PSINGLE_LIST)malloc(sizeof(SINGLE_LIST));Entry = Head1;Entry->Tag = TagIndex++;Entry->Next = NULL;for (int i = 0; i < 6; i++) {Entry->Next = (PSINGLE_LIST)malloc(sizeof(SINGLE_LIST));Entry->Next->Tag = TagIndex++;Entry->Next->Next = NULL;Entry = Entry->Next;}PSINGLE_LIST Head2 = (PSINGLE_LIST)malloc(sizeof(SINGLE_LIST));Entry = Head2;Entry->Tag = TagIndex++;Entry->Next = NULL;for (int i = 0; i < 3; i++) {Entry->Next = (PSINGLE_LIST)malloc(sizeof(SINGLE_LIST));Entry->Next->Tag = TagIndex++;Entry->Next->Next = NULL;Entry = Entry->Next;}// 构造一个交点Entry->Next = Head1->Next->Next->Next;// 下图即为此时两个链表的连接情况// +--------------------------------------------+// |   A -> B -> C -> D -> E -> F -> G          |// |                  ^                         |// |                  |                         |// |   H -> I -> J -> K                         |// +--------------------------------------------+// 方法一： // 循环遍历第一个链表中的每一项，查看是否在第二个链表中。由于两个单链表相交// 遍历当中任一链表，至少一个或多个项在第二个链表中。在最坏的情况下，将运算// m*n 次。 算法例如以下：BOOL IsIntersect = FALSE;PSINGLE_LIST IntersectEntry = NULL;Entry = Head1;while (Entry != NULL) {PSINGLE_LIST Entry2 = Head2;while (Entry2) {if (Entry == Entry2) {IsIntersect = TRUE;IntersectEntry = Entry;goto Result;}Entry2 = Entry2->Next;}Entry = Entry->Next;}Result:if (IsIntersect) {//cout << "Intersect: " << "True.   " << "Tag - " << IntersectEntry->Tag  << endl;cout << "Intersect: " << "True." << endl;} else {cout << "Intersect: " << "False" << endl;}// 方法二： // 观察上图不难发现假设两个单链表相交，那么最后一个项必定同样。所以假设站在仅仅推断单// 链表是否相交的角度来看。算法事实上能够更简单。

两个链表各遍历一遍，找到最后一个项， // 检查是否同样。运算次数为 m+n。 算法例如以下： PSINGLE_LIST LastEntry1 = NULL; PSINGLE_LIST LastEntry2 = NULL; Entry = Head1; while (Entry != NULL) { Entry = Entry->Next; } LastEntry1 = Entry; Entry = Head2; while (Entry != NULL) { Entry = Entry->Next; } LastEntry2 = Entry; if (LastEntry1 == LastEntry2) { IsIntersect = TRUE; } if (IsIntersect) { //cout << "Intersect: " << "True. " << "Tag - " << IntersectEntry->Tag << endl; cout << "Intersect: " << "True." << endl; } else { cout << "Intersect: " << "False" << endl; } }

这就是编程之美。同样的问题採用不同的方法，效果全然不同。