1.题目
相交链表：给你两个单链表的头节点 heada 和 headb ，请你找出并返回两个单链表相交的起始节点。如果两个链表没有交点，返回 null 。相交链表
2.分析
相交链表是y字型，next域相同。
定义两个引用pl和ps，
如果每个链表相交结点前长度相同，一步一步走，直到相同就找到了相交结点。如果长度不一样，首先要长链表先走差值步，然后再一人走一步直到相遇
长度不同：
长度相同：
首先求长度，先假设pl指向heada：
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listnode pl = heada; listnode ps = headb; int lena = 0; int lenb = 0; while (pl != null) { lena++; pl = pl.next; } //pl==null; pl = heada; while (ps != null) { lenb++; ps = ps.next; } //ps==null; ps = headb; |
然后根据长度差值的正负判断谁长，将pl指向长的链表：
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int len = lena - lenb;//差值步 if (len < 0) { pl = headb; ps = heada; len = lenb - lena; } |
然后长的先走长度差值步，最后一人一步走：
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//pl走差值len步 while (len != 0) { pl = pl.next; len--; } //同时走，直到相遇 while (pl != ps) { pl = pl.next; ps = ps.next; } return pl; } |
3.完整代码
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//判断链表相交 public static listnode getintersectionnode(listnode heada, listnode headb) { if (heada == null || headb == null) { return null; } listnode pl = heada; listnode ps = headb; int lena = 0; int lenb = 0; while (pl != null) { lena++; pl = pl.next; } //pl==null; pl = heada; while (ps != null) { lenb++; ps = ps.next; } //ps==null; ps = headb; int len = lena - lenb;//差值步 if (len < 0) { pl = headb; ps = heada; len = lenb - lena; } //1、pl永远指向最长的链表 ps永远指向最短的链表 2、求到了差值len步 //pl走差值len步 while (len != 0) { pl = pl.next; len--; } //同时走，直到相遇 while (pl != ps) { pl = pl.next; ps = ps.next; } return pl; } |
测试：
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public static void main(string[] args) { mylinkedlist mylinkedlist = new mylinkedlist(); mylinkedlist.addlast(12); mylinkedlist.addlast(23); mylinkedlist.addlast(34); mylinkedlist.addlast(45); system.out.println("mylinkedlist："); mylinkedlist.display(); mylinkedlist mylinkedlist1 = new mylinkedlist(); mylinkedlist1.addlast(13); mylinkedlist1.addlast(22); mylinkedlist1.addlast(30); system.out.println("mylinkedlist1："); mylinkedlist1.display(); createcut(mylinkedlist.head, mylinkedlist1.head); try { listnode ret = getintersectionnode(mylinkedlist.head, mylinkedlist1.head); mylinkedlist.display2(ret); } catch (nullpointerexception e) { e.printstacktrace(); system.out.println("没有相交结点！"); } } |
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mylinkedlist mylinkedlist = new mylinkedlist(); mylinkedlist.addlast(12); mylinkedlist.addlast(23); mylinkedlist.addlast(34); mylinkedlist.addlast(56); system.out.println("mylinkedlist:"); mylinkedlist.display(); mylinkedlist mylinkedlist1 = new mylinkedlist(); mylinkedlist1.addlast(12); mylinkedlist1.addlast(23); mylinkedlist1.addlast(30); system.out.println("mylinkedlist1:"); mylinkedlist1.display(); //createcut(mylinkedlist.head,mylinkedlist1.head); try { listnode ret = getintersectionnode(mylinkedlist.head, mylinkedlist1.head); system.out.println(ret.val); }catch (nullpointerexception e){ e.printstacktrace(); system.out.println("不存在相交结点！"); } } |
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原文链接：https://blog.csdn.net/qq_44721738/article/details/121190579
