算法练习_LeetCode_链表1

最近有空就在LeetCode上刷下题，在工作中虽然很少自已写算法，JDK已经的封装好了，直接拿来就用，但是平常有空刷下题对于理解这些封装的集合API及选用还是挺有帮助的。毕竟不是计算机科班出身的，数据结构与算法平常还是要多补下。

• 删除链表倒数第n个数
  

Given a linked list, remove the nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        //计算长度        ListNode point = head;        int length = 1;        while (null != point.next) {            point = point.next;            length++;        }        if (n > length)        {            return head;        }        ListNode temp = head;        if (length == n) {            head = temp.next;            temp.next = null;            return head;        }        int i = 1;        while (null != temp.next && i + n < length) {            temp = temp.next;            i++;        }        //此时temp处于倒数n+1个node上        ListNode nthNode = temp.next;        temp.next = nthNode.next;        nthNode.next = null;        return head;    }}

• 删除链表中相同的元素
  

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

public ListNode deleteDuplicates(ListNode head) {    //只有一个节点或节点为空    if (null == head || null == head.next) {        return head;    }    ListNode point = head;    ListNode temp = null;    int flag = 0;    while (null != point && null != (temp = point.next)) {        //前N个节点相同的情况        while (null != temp && point.val == temp.val) {            temp = temp.next;            point = point.next;            flag++;        }        if (0 != flag) {            point.next = null;            point = temp;            head = point;  //表头重新赋值            flag = 0;            continue;        }        while (null!= temp.next && temp.val == temp.next.val) {            temp = temp.next;            flag++;        }        if (0 != flag) {            point.next = temp.next;            temp.next = null;            flag = 0;            continue;   //链表重新组装后需要再次进入循环        }        point = point.next;    }    return head;}

• 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}.

public void reorderList(ListNode head) {     //排除null，节点数为1或2的List    if (null == head || null == head.next || null == head.next.next) {        return;    }ListNode point = head;    ListNode secondaryEnd = head.next;  //倒数第二个节点    while (null != secondaryEnd.next && null != secondaryEnd.next.next) {        secondaryEnd = secondaryEnd.next;    }    ListNode end = secondaryEnd.next;   //最后一个节点    if (null != point.next) {        secondaryEnd.next = null;        end.next = point.next;        point.next = end;        point = end.next;        reorderList(point);    }   }

这种解法的平均时间复杂度应该是O(n2) , 在LeetCode中报超时，暂时还未想到更快的办法。