判断两条单链表是否相交

来源：互联网发布：linux 文件排序编辑：程序博客网时间：2024/06/06 00:54

算法中经常会判断两条单链表是否相交，虽然算法简单，但也值得说一下。代码中有详尽注释， 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，因为两个单链表相交    // 遍历其中任一链表，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。 算法如下：    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：     // 观察上图不难发现如果两个单链表相交，那么最后一个项必然相同。所以如果站在只判断单    // 链表是否相交的角度来看，算法其实可以更简单。两个链表各遍历一遍，找到最后一个项，    // 检查是否相同。运算次数为 m+n。 算法如下：    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，相同的问题采用不同的方法，效果完全不同。

133 0