玩转数据结构——链表和递归

递归

本质上，将原来的问题，转化为更小的同一问题

举例：数组求和

 1 /*
 2 
 3     Sum(arr[0...n-1]) = arr[0] + Sum(arr[1...n-1]) <-- 更小的同一问题
 4     Sum(arr[1...n-1]) = arr[1] + Sum(arr[2...n-1]) <-- 更小的同一问题
 5 
 6                             ......
 7     Sum(arr[n-1...n-1]) = arr[n-1] + Sum(arr[]) <-- 更小的同一问题
 8 
 9 */
10 
11 public class Sum {
12 
13     public static int sum(int[] arr){
14         return sum(arr, 0);
15     }
16 
17     // 计算arr[l...n)这个区间内所有数字的和
18     private static int sum(int[] arr, int l){
19         if(l == arr.length)      
20             return 0;　　　　// <-- 求解最基本问题　　　　
21         return arr[l] + sum(arr, l + 1);　　// <-- 把原问题转化成更小的问题
22     }
23 
24     public static void main(String[] args) {
25 
26         int[] nums = {1, 2, 3, 4, 5, 6, 7, 8};
27         System.out.println(sum(nums));
28     }
29 }

注意递归函数的“宏观”语意

递归函数就是一个函数，完成一个功能

Definition for singly-linked list. 1 public class ListNode {  3 public int val; 4 public ListNode next 5 6 public ListNode(int x) {

 7 val = x;  8  }  9 10 // 链表节点的构造函数 11 // 使用arr为参数，创建一个链表，当前的ListNode为链表头结点 12 public ListNode(int[] arr){ 13 14 if(arr == null || arr.length == 0) 15 throw new IllegalArgumentException("arr can not be empty"); 16 17 this.val = arr[0]; 18 ListNode cur = this; 19 for(int i = 1 ; i < arr.length ; i ++){ 20 cur.next = new ListNode(arr[i]); 21 cur = cur.next; 22  } 23  } 24 25 // 以当前节点为头结点的链表信息字符串 26  @Override 27 public String toString(){ 28 29 StringBuilder s = new StringBuilder(); 30 ListNode cur = this; 31 while(cur != null){ 32 s.append(cur.val + "->"); 33 cur = cur.next; 34  }

　　　　　　　　/*
　　　　　　　　for(cur = this; current != NULL; cur = cur.next){
　　　　　　　　　　s.append(cur.val + "-->");
　　　　　　　　}
　　　　　　　　*/

35         s.append("NULL");
36         return s.toString();
37     }
38 }

 1 class Solution {
 2 
 3     public ListNode removeElements(ListNode head, int val) {
 4 
 5         while(head != null && head.val == val){
 6             ListNode delNode = head;
 7             head = head.next;
 8             delNode.next = null;
 9         }
10 
11         if(head == null)
12             return head;
13 
14         ListNode prev = head;
15         while(prev.next != null){
16             if(prev.next.val == val) {
17                 ListNode delNode = prev.next;
18                 prev.next = delNode.next;
19                 delNode.next = null;
20             }
21             else
22                 prev = prev.next;
23         }
24 
25         return head;
26     }
27 
28     public static void main(String[] args) {
29 
30         int[] nums = {1, 2, 6, 3, 4, 5, 6};
31         ListNode head = new ListNode(nums);
32         System.out.println(head);
33 
34         ListNode res = (new Solution()).removeElements(head, 6);
35         System.out.println(res);
36     }
37 }

 1 /// Leetcode 203. Remove Linked List Elements
 2 /// https://leetcode.com/problems/remove-linked-list-elements/description/
 3 
 4 class Solution2 {
 5 
 6     public ListNode removeElements(ListNode head, int val) {
 7 
 8         while(head != null && head.val == val)
 9             head = head.next;
10 
11         if(head == null)
12             return head;
13 
14         ListNode prev = head;
15         while(prev.next != null){
16             if(prev.next.val == val)
17                 prev.next = prev.next.next;
18             else
19                 prev = prev.next;
20         }
21 
22         return head;
23     }
24 
25     public static void main(String[] args) {
26 
27         int[] nums = {1, 2, 6, 3, 4, 5, 6};
28         ListNode head = new ListNode(nums);
29         System.out.println(head);
30 
31         ListNode res = (new Solution2()).removeElements(head, 6);
32         System.out.println(res);
33     }
34 }

Solution2

 1 class Solution3 {
 2 
 3     public ListNode removeElements(ListNode head, int val) {
 4 
 5         ListNode dummyHead = new ListNode(-1);
 6         dummyHead.next = head;
 7 
 8         ListNode prev = dummyHead;
 9         while(prev.next != null){
10             if(prev.next.val == val)
11                 prev.next = prev.next.next;
12             else
13                 prev = prev.next;
14         }
15 
16         return dummyHead.next;
17     }
18 
19     public static void main(String[] args) {
20 
21         int[] nums = {1, 2, 6, 3, 4, 5, 6};
22         ListNode head = new ListNode(nums);
23         System.out.println(head);
24 
25         ListNode res = (new Solution3()).removeElements(head, 6);
26         System.out.println(res);
27     }
28 }

Solution3

 1 class Solution4 {
 2 
 3     public ListNode removeElements(ListNode head, int val) {
 4 
 5         if(head == null)
 6             return head;
 7 
 8         ListNode res = removeElements(head.next, val);
 9         if(head.val == val)
10             return res;
11         else{
12             head.next = res;
13             return head;
14         }
15     }
16 
17     public static void main(String[] args) {
18 
19         int[] nums = {1, 2, 6, 3, 4, 5, 6};
20         ListNode head = new ListNode(nums);
21         System.out.println(head);
22 
23         ListNode res = (new Solution4()).removeElements(head, 6);
24         System.out.println(res);
25     }
26 }

Solution4

 1 class Solution5 {
 2 
 3     public ListNode removeElements(ListNode head, int val) {
 4 
 5         if(head == null)
 6             return head;
 7 
 8         head.next = removeElements(head.next, val);
 9         return head.val == val ? head.next : head;
10     }
11 
12     public static void main(String[] args) {
13 
14         int[] nums = {1, 2, 6, 3, 4, 5, 6};
15         ListNode head = new ListNode(nums);
16         System.out.println(head);
17 
18         ListNode res = (new Solution5()).removeElements(head, 6);
19         System.out.println(res);
20     }
21 }

Solution5