多个有序链表的合并

1, 先将问题简化,合并两个有序链表

首先分析合并两个链表的过程。我们的分析从合并两个链表的头结点开始。链表1的头结点的值小于链表2的头结点的值，因此链表1的头结点将是合并后链表的头结点。如下图所示。

参考:http://www.cnblogs.com/jason2013/articles/4341153.html

使用递归方法,一步步生成头结点,代码如下

递归的要诀是子问题要和父问题完全一样,只是规模变小(每次调用,更小的参数值),

 1 List merge(List head1, List head2){
 2     List mergeHead = NULL;
 3     if (head1 == NULL) {
 4         return head2;
 5     }
 6     if (head2 == NULL){
 7         return head1;
 8     }
 9
10     if (head1->item < head2->item){
11         mergeHead = head1;
12         mergeHead->next = merge(head1->next, head2);
13     }else{
14         mergeHead = head2;
15         mergeHead->next = merge(head1, head2->next);
16     }
17     return mergeHead;
18 }

2, 当有多个链表时,考虑分治法每两个链表进行合并.

确定父子函数原型,要想使用递归,子问题和父问题必须完全一样(即返回值,参数类型完全一致)

父问题:多个链表

子问题:n/2,...,2个链表,1个链表

递归函数原型List mergeList(int l, int r)

父子问题都是返回一个合并后链表,

使用l,r 两个变量控制问题规模,指定链表个数(快速排序,归并排序都喜欢用这样的两个参数)

将多个链表存放在全局变量vector<List> lists中,简化递归函数.

第9行代码复用前面提到的两个有序链表合并

 1 List mergeList(int l, int r){
 2     List u, v;
 3     int m = (l + r) / 2;
 4     if (l == r) {
 5         return lists[l];
 6     }
 7     u = mergeList(l, m);
 8     v = mergeList(m + 1, r);
 9     return merge(u, v);
10 }

3, main 函数

 1 int main(void)
 2 {
 3     int size = 8;
 4     int num = 5;
 5     ListFactory(size, num);
 6     for (int i = 0; i < size; i++){
 7         print(lists[i]);
 8     }
 9     cout << endl;
10     link t = mergeList(0, size-1);
11     print(t);
12     return 0;
13 }

效果

1->9->17->25->33->
2->10->18->26->34->
3->11->19->27->35->
4->12->20->28->36->
5->13->21->29->37->
6->14->22->30->38->
7->15->23->31->39->
8->16->24->32->40->

1->2->3->4->5->6->7->8->9->10->11->12->13->14->15->16->17->18->19->20->21->22->23->24->25->26->27->28->29->30->31->32->33->34->35->36->37->38->39->40->

完整程序

 1 #include<iostream>
 2 #include<string>
 3 #include<vector>
 4 using std::cin;
 5 using std::cout;
 6 using std::endl;
 7 using std::string;
 8 using std::vector;
 9 typedef struct node* link;
10 struct node{
11     int item;
12     link next;
13 };
14 typedef link List;
15 vector<List> lists;
16 void print(List list){
17     while (list != NULL){
18         cout << list->item<< "->";
19         list = list->next;
20     }
21     cout << endl;
22 }
23
24 vector<List> ListFactory(int num, int size){
25     for (int k = 1; k <= num; k++){
26         link t = (link)malloc(sizeof *t);
27         t->item = k;
28         t->next = t;
29         link x = t;
30         for (int m = k + num; m <= num*size; m = m+num){
31             x = (x->next = (link)malloc(sizeof *x));
32             x->item = m;
33             x->next = t;
34         }
35         x->next = NULL;
36         lists.push_back(t);
37     }
38     return lists;
39 }
40
41 List merge(List head1, List head2){
42     List mergeHead = NULL;
43     if (head1 == NULL) {
44         return head2;
45     }
46     if (head2 == NULL){
47         return head1;
48     }
49
50     if (head1->item < head2->item){
51         mergeHead = head1;
52         mergeHead->next = merge(head1->next, head2);
53     }else{
54         mergeHead = head2;
55         mergeHead->next = merge(head1, head2->next);
56     }
57     return mergeHead;
58 }
59
60 List mergeList(int l, int r){
61     List u, v;
62     int m = (l + r) / 2;
63     if (l == r) {
64         return lists[l];
65     }
66     u = mergeList(l, m);
67     v = mergeList(m + 1, r);
68     return merge(u, v);
69 }
70
71 int main(void)
72 {
73     int size = 8;
74     int num = 5;
75     ListFactory(size, num);
76     for (int i = 0; i < size; i++){
77         print(lists[i]);
78     }
79     cout << endl;
80     link t = mergeList(0, size-1);
81     print(t);
82     return 0;
83 }