List综合应用

148. Sort List

Sort a linked list in O(n log n) time using constant space complexity.

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } *///第一种方法，mergeSort//①找中点//②把两边排好序//③merge两个有序数组// class Solution {//     //寻找中点，通过快慢指针实现//     public ListNode findMiddle(ListNode head){//         ListNode fast = head.next,slow = head;//         while(fast!=null && fast.next!=null){//             slow = slow.next;//             fast = fast.next.next;//         }//         return slow;//     }    //     public ListNode merge(ListNode left,ListNode right){//         ListNode dummpy = new ListNode(0);//         ListNode lastNode = dummpy;//         while(left != null && right != null){//             if(left.val < right.val){//                 lastNode.next = left;//                 left = left.next;//             }else{//                 lastNode.next = right;//                 right = right.next;//             }//             lastNode = lastNode.next;//         }//         if(left != null)//             lastNode.next = left;//         if(right != null)//             lastNode.next = right;//         return dummpy.next;//     }//     public ListNode sortList(ListNode head) {//         if(head == null || head.next == null) return head;//如果不加head.next==null会陷入死循环//         ListNode middle = findMiddle(head);//         ListNode left = sortList(middle.next);//         middle.next = null;//         ListNode right = sortList(head);//         return merge(left,right);//     }// }//第二种方法，quickSort//①找中点//②根据中点partition List//③对两个链表分别排序//④连接链表class Solution {    //寻找中点，通过快慢指针实现    public ListNode findMiddle(ListNode head){        ListNode fast = head.next,slow = head;        while(fast!=null && fast.next!=null){            slow = slow.next;            fast = fast.next.next;        }        return slow;    }    public ListNode getTail(ListNode head){        if(head == null) return head;        while(head.next!= null)            head = head.next;          return head;    }    public ListNode concat(ListNode left,ListNode middle,ListNode right){        ListNode dummpy = new ListNode(0), tail = dummpy;        tail.next = left;tail = getTail(tail);        tail.next = middle;tail = getTail(tail);        tail.next = right;        return dummpy.next;    }        public ListNode sortList(ListNode head) {        if(head == null || head.next == null) return head;//如果不加head.next==null会陷入死循环                ListNode middle = findMiddle(head);                //partition链表        ListNode leftDummpy = new ListNode(0),leftTail = leftDummpy;        ListNode rightDummpy = new ListNode(0),rightTail = rightDummpy;        ListNode middleDummy = new ListNode(0), middleTail = middleDummy;        while(head != null){            if(head.val < middle.val){                leftTail.next = head;                leftTail = head;            }else if(head.val > middle.val){                rightTail.next = head;                rightTail = head;            }else{                middleTail.next = head;                middleTail = head;            }            head = head.next;        }                leftTail.next = null;        middleTail.next = null;        rightTail.next = null;                    ListNode left = sortList(leftDummpy.next);        ListNode right = sortList(rightDummpy.next);        return concat(left,middleDummy.next,right);    }}

143. Reorder List

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } *///寻找中点//根据中点把链表分成两半，并且逆转右边//merge两个左右链表class Solution {    public void reorderList(ListNode head) {        if(head == null || head.next == null) return ;        ListNode middle = findMiddle(head);        ListNode tail = reverse(middle.next);        middle.next = null;        merge(head,tail);    }    public ListNode findMiddle(ListNode head){        ListNode slow = head,fast = head.next;        while(fast != null && fast.next != null){            slow = slow.next;            fast = fast.next.next;        }        return slow;    }    public ListNode reverse(ListNode head){        ListNode pre = null;        while(head != null){            ListNode temp = head.next;            head.next = pre;            pre = head;            head = temp;        }        return pre;    }    public void merge(ListNode head,ListNode tail){        ListNode dummpy = new ListNode(0);        int index = 0;        while(head != null && tail != null){            if(index % 2 == 0){                dummpy.next = head;                head = head.next;            }else{                dummpy.next = tail;                tail = tail.next;            }            index++;            dummpy = dummpy.next;        }        if(head != null)            dummpy.next = head;        else            dummpy.next = tail;    }}