复杂链表复制

题目

复杂链表的复制。一个链表的每个节点，有一个指向next指针指向下一个节点，还有一个random指针指向这个链表中的一个随机节点或者NULL，现在要求实现复制这个链表，返回复制后的新链表。

解法一

（暴力求解）
1. 这个复杂链表是以单链表为基础的，可以先复制一个单链表，random 指针在原链表上的位置与新链表是相对应的，建立同步投影指针 shadow，与 cur 在原链表上同步，找新链表上的 random ，（如果找不到，说明原链表指针指向了未知地址）。
2. 考虑复杂情况，原链表带环，可以找到入口点并记录，将带环链表转化为无环链表处理，再处理无环的情况。

写代码

1. 处理空链表情况
2. 用找环入口点函数判断是否带环
3. 若带环，先复制尾，到入口点，复制环，又到入口点，复制完毕，拆环（原链表），保留尾，接环用。然后就是不带环链表 random 指针的复制。
4. 若不带环，不带环链表复制。
5. 复制 random 封装函数（要用两次，还很长）。

代码

//尾插void PushBackCom(ComplexNode** ppList, int x){    assert(ppList);    ComplexNode* newTail = (ComplexNode*)malloc(sizeof(ComplexNode));    newTail->data = x;    newTail->next = NULL;    newTail->random = NULL;    if (*ppList == NULL)        *ppList = newTail;    else{        ComplexNode* plist = *ppList;        while (plist->next != NULL){            plist = plist->next;        }        plist->next = newTail;    }}//求入口点ComplexNode* IntranceOfRingCom(ComplexNode* pList){    ComplexNode* fast = pList;    ComplexNode* slow = pList;    while (fast && fast->next)    {        fast = fast->next->next;        slow = slow->next;        if (fast == slow)            break;    }    if (fast != slow || fast->next == NULL)        return NULL;    while (pList != slow)    {        pList = pList->next;        slow = slow->next;    }    return pList;}void copyRandom(ComplexNode* cur, ComplexNode* n_cur){    ComplexNode* newComList = n_cur;    ComplexNode* pList = cur;    while (n_cur != NULL)    {        if (cur->random != NULL)        {            ComplexNode* shadow = newComList;            ComplexNode* cur2 = pList;            while (cur->random != cur2 && cur2 != NULL)            {                shadow = shadow->next;                cur2 = cur2->next;            }            n_cur->random = shadow;            if (shadow == NULL)//在这里出现NULL，意味着有random指向非本链表的位置                printf("warn : point to unknowed node.");        }        else            n_cur->random = NULL;        n_cur = n_cur->next;        cur = cur->next;    } }ComplexNode* copyComplexList(ComplexNode* pList){    if (pList == NULL)return NULL;    ComplexNode* intr = IntranceOfRingCom(pList);    ComplexNode* newComList = NULL;    PushBackCom(&newComList, pList->data);    if (intr != NULL)    {        ComplexNode* n_intr = newComList;         ComplexNode* n_cur = newComList;        ComplexNode* cur = pList;        while (cur != intr)//尾复制        {            cur = cur->next;            PushBackCom(&n_cur, cur->data);            n_cur = n_cur->next;            n_intr = n_intr->next;        }        while (cur->next != intr)//环复制        {            cur = cur->next;            PushBackCom(&n_cur, cur->data);            n_cur = n_cur->next;        }        copyRandom(pList, newComList);        //接环，现在n_cur刚好是尾        n_cur->next = n_intr;    }    else    {        ComplexNode* n_cur = newComList;        ComplexNode* cur = pList;        while (cur->next != NULL)        {            cur = cur->next;            PushBackCom(&n_cur, cur->data);            n_cur = n_cur->next;        }        copyRandom(pList, newComList);    }    return newComList;}
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
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
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
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109

方法二

（没有处理带环情况，可参考方法一处理带环情况）

代码

ComplexNode* copyComplexList2(ComplexNode* pList){    if (pList == NULL)return NULL;    ComplexNode* cur = pList;    ComplexNode* newcode;    while (cur != NULL)    {        newcode = (ComplexNode*)malloc(sizeof(ComplexNode));        newcode->data = cur->data;        newcode->next = cur->next;        newcode->random = NULL;        cur->next = newcode;        cur = cur->next->next;    }    cur = pList;    newcode = pList->next;    while (cur != NULL)    {        if (cur->random != NULL)            newcode->random = cur->random->next;        newcode = cur->next;        cur = cur->next->next;    }    newcode = pList->next;    ComplexNode* ret = pList->next;    cur = pList;    cur->next = cur->next->next;    cur = cur->next;    while (cur != NULL)    {        newcode->next = newcode->next->next;        cur->next = cur->next->next;        cur = cur->next;        newcode = newcode->next;    }    return ret;}
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
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

(function () {('pre.prettyprint code').each(function () { var lines = (this).text().split(′\n′).length;varnumbering = $('

').addClass('pre-numbering').hide(); (this).addClass(′has−numbering′).parent().append(numbering); for (i = 1; i