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