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這篇文章主要為大家展示了“java如何實(shí)現(xiàn)相交鏈表”,內(nèi)容簡(jiǎn)而易懂,條理清晰,希望能夠幫助大家解決疑惑,下面讓小編帶領(lǐng)大家一起研究并學(xué)習(xí)一下“java如何實(shí)現(xiàn)相交鏈表”這篇文章吧。
相交鏈表:給你兩個(gè)單鏈表的頭節(jié)點(diǎn) headA 和 headB ,請(qǐng)你找出并返回兩個(gè)單鏈表相交的起始節(jié)點(diǎn)。如果兩個(gè)鏈表沒(méi)有交點(diǎn),返回 null 。相交鏈表
相交鏈表是Y字型
,next域
相同。
定義兩個(gè)引用pl和ps
,
如果每個(gè)鏈表相交結(jié)點(diǎn)前長(zhǎng)度相同,一步一步走,直到相同就找到了相交結(jié)點(diǎn)。如果長(zhǎng)度不一樣,首先要長(zhǎng)鏈表先走差值步,然后再一人走一步直到相遇
長(zhǎng)度不同:
長(zhǎng)度相同:
首先求長(zhǎng)度,先假設(shè)pl指向headA:
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;
然后根據(jù)長(zhǎng)度差值的正負(fù)判斷誰(shuí)長(zhǎng),將pl指向長(zhǎng)的鏈表:
int len = lenA - lenB;//差值步 if (len < 0) { pl = headB; ps = headA; len = lenB - lenA; }
然后長(zhǎng)的先走長(zhǎng)度差值步,最后一人一步走:
//pl走差值len步 while (len != 0) { pl = pl.next; len--; } //同時(shí)走,直到相遇 while (pl != ps) { pl = pl.next; ps = ps.next; } return pl; }
//判斷鏈表相交 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永遠(yuǎn)指向最長(zhǎng)的鏈表 ps永遠(yuǎn)指向最短的鏈表 2、求到了差值len步 //pl走差值len步 while (len != 0) { pl = pl.next; len--; } //同時(shí)走,直到相遇 while (pl != ps) { pl = pl.next; ps = ps.next; } return pl; }
測(cè)試:
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("沒(méi)有相交結(jié)點(diǎn)!"); } }
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("不存在相交結(jié)點(diǎn)!"); } }
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