[算法]頭條面試—奇數位升序偶數位降序鏈表排序

題目描述：一個鏈表，奇數位升序偶數位降序，讓鏈表變成升序的。

好比：1 8 3 6 5 4 7 2 9，最後輸出1 2 3 4 5 6 7 8 9。

分析：

這道題能夠分紅三步：

首先根據奇數位和偶數位拆分紅兩個鏈表。

而後對偶數鏈表進行反轉。

最後將兩個有序鏈表進行合併。

package com.darrenchan.list; public class OddIncreaseEvenDecrease { /** * 按照奇偶位拆分紅兩個鏈表 * @param head * @return
     */
    public static Node[] getLists(Node head){ Node head1 = null; Node head2 = null; Node cur1 = null; Node cur2 = null; int count = 1;//用來計數
        while(head != null){ if(count % 2 == 1){ if(cur1 != null){ cur1.next = head; cur1 = cur1.next; }else{ cur1 = head; head1 = cur1; } }else{ if(cur2 != null){ cur2.next = head; cur2 = cur2.next; }else{ cur2 = head; head2 = cur2; } } head = head.next; count++; } //跳出循環，要讓最後兩個末尾元素的下一個都指向null
        cur1.next = null; cur2.next = null; Node[] nodes = new Node[]{head1, head2}; return nodes; } /** * 反轉鏈表 * @param head * @return
     */
    public static Node reverseList(Node head){ Node pre = null; Node next = null; while(head != null){ next = head.next; head.next = pre; pre = head; head = next; } return pre; } /** * 合併兩個有序鏈表 * @param head1 * @param head2 * @return
     */
    public static Node CombineList(Node head1, Node head2){ if(head1 == null || head2 == null){ return head1 != null ? head1 : head2; } Node head = head1.value < head2.value ? head1 : head2; Node cur1 = head == head1 ? head1 : head2; Node cur2 = head == head1 ? head2 : head1; Node pre = null; Node next = null; while(cur1 != null && cur2 != null){ if(cur1.value <= cur2.value){//這裏必定要有=，不然一旦cur1的value和cur2的value相等的話，下面的pre.next會出現空指針異常
                pre = cur1; cur1 = cur1.next; }else{ next = cur2.next; pre.next = cur2; cur2.next = cur1; pre = cur2; cur2 = next; } } pre.next = cur1 == null ? cur2 : cur1; return head; } public static void main(String[] args) { Node head = init(); Node[] lists = getLists(head); Node head1 = lists[0]; Node head2 = lists[1]; head2 = reverseList(head2); head = CombineList(head1, head2); while(head != null){ System.out.println(head.value); head = head.next; } } public static Node init(){ Node node1 = new Node(1); Node node2 = new Node(8); Node node3 = new Node(3); Node node4 = new Node(6); Node node5 = new Node(5); Node node6 = new Node(4); Node node7 = new Node(7); Node node8 = new Node(2); Node node9 = new Node(9); node1.next = node2; node2.next = node3; node3.next = node4; node4.next = node5; node5.next = node6; node6.next = node7; node7.next = node8; node8.next = node9; return node1; } }

 

合併兩個單鏈表，還能夠採用遞歸的方式：

public class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if(l1 == null && l2 == null) return null; if(l1 == null) return l2; if(l2 == null) return l1; ListNode head = null; if(l1.val > l2.val) { head = l2; head.next = mergeTwoLists(l1, l2.next); } else { head = l1; head.next = mergeTwoLists(l1.next, l2); } return head; } }