java实现链表基本操作

在看剑指offer的时候看到有关链表基本操作的考点，想总结一下，基本是参考http://blog.csdn.net/kerryfish/article/details/24043099文章实现的，对于该文章中的方法进行了编写，测试，修改了其中几个问题。源码如下

package cn.com.cnn;import java.util.Stack;public class ListSum {    public static class Node {        int value;        Node next;        public Node(int value) {            this.next = null;            this.value = value;        }    }    public static void main(String[] args) {        int[] data1 = {1, 2, 3, 4, 5};        int[] data2 = {3, 4, 5};        ListSum listSum = new ListSum();        Node pHead = listSum.createList(data1);        Node merge = listSum.createList(data2);        //listSum.printList(pHead);        //System.out.println(listSum.reGetKthNode(pHead, 3));        //listSum.printList(listSum.reverseList(pHead));        //listSum.printList(listSum.reverseListRec(pHead));        //listSum.returnMiddNode(pHead);        //listSum.reversePrintListRec(pHead);        //listSum.reversePrintListStack(pHead);        //listSum.printList(listSum.mergeLists(pHead, merge));        //listSum.printList(listSum.deleteNode(pHead, 2));        //listSum.printList(listSum.firstCommonNode(pHead, merge));    }    public Node createList(int[] data) {        Node pHead = null;        Node temp = null;        for(int i = 0; i < data.length; i++) {            if(i == 0) {                pHead = new Node(data[i]);                temp = pHead;                continue;            }            temp.next = new Node(data[i]);            temp = temp.next;        }        return pHead;    }    public static int getListLength(Node pHead) {        int length = 0;        while(pHead != null) {            length++;            pHead = pHead.next;        }        return length;    }    //利用循环倒置链表    public Node reverseList(Node pHead) {        if(pHead == null || pHead.next == null ) {            return pHead;        }        Node pNext = null;        Node pPre = null;        while(pHead != null) {            pNext = pHead.next;            pHead.next = pPre;            pPre = pHead;            pHead = pNext;        }        return pPre;    }    //利用迭代倒置链表     public Node reverseListRec(Node pHead) {        if(pHead == null || pHead.next == null) {            return pHead;        }        Node pNext = reverseListRec(pHead.next);        pHead.next = pNext;        pHead = pHead.next;        return pNext;    }    //查找链表中间节点，两个指针一个一次跳两个，一个一次跳一个    public Node returnMiddNode(Node pHead) {        if(pHead == null || pHead.next == null) {            return pHead;        }        Node pMidd = pHead;        Node temp = pHead;        while(temp != null && temp.next != null) {            temp = temp.next.next;            pMidd = pMidd.next;        }        System.out.println(pMidd.value);        return pMidd;    }    public int reGetKthNode(Node pHead, int k) {        Node p = pHead;        if(pHead == null || k <= 0) {            return 0;        }        for(int i = 0; i < k; i++) {            p = p.next;        }        while(p.next != null) {            p = p.next;            pHead = pHead.next;        }        return pHead.value;     }    public void printList(Node pHead) {        if(pHead == null) {            System.out.println("null");        }        while(pHead != null) {            System.out.print(pHead.value + ", ");            pHead = pHead.next;        }    }    //从尾到头打印单链表，递归    public void reversePrintListRec(Node pHead) {        if(pHead == null) {            return;        } else {            reversePrintListRec(pHead.next);            System.out.print(pHead.value + " ");        }    }    //从尾到头打印单链表，栈    public void reversePrintListStack(Node pHead) {        Stack<Node> nodeStack = new Stack<Node>();        while(pHead != null) {            nodeStack.push(pHead);            pHead = pHead.next;        }        while(!nodeStack.empty()) {            System.out.print(nodeStack.pop().value + " ");        }    }    //合并两个有序的单链表l1,l2     public Node mergeLists(Node head1, Node head2) {        if(head1 == null) {            return head2;        } else if(head2 == null) {            return head1;        }        Node result = null;        Node temp = null;        if(head1.value < head2.value) {            temp = head1;            head1 = head1.next;        } else if(head1.value > head2.value) {            temp = head2;            head2= head2.next;        } else {            temp = head1;            head1 = head1.next;            head2 = head2.next;        }        result = temp;        while(head1 != null && head2 != null) {            if(head1.value < head2.value) {                temp.next =head1;                head1 = head1.next;            } else if(head1.value > head2.value) {                temp.next = head1;                head1= head1.next;            } else {                temp.next = head1;                head1 = head1.next;                head2 = head2.next;            }            temp = temp.next;            temp.next = null;        }        if(head1 == null) {            temp.next = head2;        }        if(head2 == null) {            temp.next = head1;        }        return result;    }    public Node deleteNode(Node pHead, int value) {        Node temp = null;        if(pHead == null) {            return temp;        }        temp = pHead;        if(temp.value == value) {            temp = temp.next;            pHead = temp;        } else {            while(temp.next != null && temp.next.value != value) {                temp = temp.next;            }            if(temp.next != null && temp.next.value == value) {                temp.next = temp.next.next;            }        }        return pHead;    }    //求两个链表的第一个公共节点    //如果仅仅判断两根链表是否交叉，直接看两个链表最后一个节点是不是相等就可以了。因为都是单链表    public Node firstCommonNode(Node list1, Node list2) {        Node result = null;        if(list1 == null || list2 == null) {            return result;        }        int list1Length = getListLength(list1);        int list2Length = getListLength(list2);        if(list1Length <= list2Length) {            for(int i = 0; i < list2Length - list1Length; i++) {                list2 = list2.next;            }        } else {            for(int i = 0; i < list1Length - list2Length; i++) {                list1 =list1.next;            }        }        while(list1 != null && list2 != null) {            if(list1.value == list2.value) {                result = list1;                result.next = null;                break;            } else {                list1 = list1.next;                list2 = list2.next;            }        }        return result;    }    //给出head,delete指针，O(1)时间复杂度删除delete。将删除节点和下个节点的值置换再删除    public Node deleteNode2(Node pHead, Node delete) {        Node result = null;        if(delete == null) {            return result;        }        if(delete.next == null) {            if(pHead == delete) {                return result;            } else {                Node temp = pHead;                while(temp.next != delete) {                    temp = temp.next;                }                temp.next = null;                return pHead;            }        }        delete.value = delete.next.value;        delete.next = delete.next.next;        return result;    }}