-
Notifications
You must be signed in to change notification settings - Fork 0
/
138.copy-list-with-random-pointer.cpp
139 lines (135 loc) · 3.41 KB
/
138.copy-list-with-random-pointer.cpp
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
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
/*
* @lc app=leetcode id=138 lang=cpp
*
* [138] Copy List with Random Pointer
*
* https://leetcode.com/problems/copy-list-with-random-pointer/description/
*
* algorithms
* Medium (39.23%)
* Likes: 4592
* Dislikes: 789
* Total Accepted: 516.3K
* Total Submissions: 1.3M
* Testcase Example: '[[7,null],[13,0],[11,4],[10,2],[1,0]]'
*
* A linked list of length n is given such that each node contains an
* additional random pointer, which could point to any node in the list, or
* null.
*
* Construct a deep copy of the list. The deep copy should consist of exactly n
* brand new nodes, where each new node has its value set to the value of its
* corresponding original node. Both the next and random pointer of the new
* nodes should point to new nodes in the copied list such that the pointers in
* the original list and copied list represent the same list state. None of the
* pointers in the new list should point to nodes in the original list.
*
* For example, if there are two nodes X and Y in the original list, where
* X.random --> Y, then for the corresponding two nodes x and y in the copied
* list, x.random --> y.
*
* Return the head of the copied linked list.
*
* The linked list is represented in the input/output as a list of n nodes.
* Each node is represented as a pair of [val, random_index] where:
*
*
* val: an integer representing Node.val
* random_index: the index of the node (range from 0 to n-1) that the random
* pointer points to, or null if it does not point to any node.
*
*
* Your code will only be given the head of the original linked list.
*
*
* Example 1:
*
*
* Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
* Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]
*
*
* Example 2:
*
*
* Input: head = [[1,1],[2,1]]
* Output: [[1,1],[2,1]]
*
*
* Example 3:
*
*
*
*
* Input: head = [[3,null],[3,0],[3,null]]
* Output: [[3,null],[3,0],[3,null]]
*
*
* Example 4:
*
*
* Input: head = []
* Output: []
* Explanation: The given linked list is empty (null pointer), so return
* null.
*
*
*
* Constraints:
*
*
* 0 <= n <= 1000
* -10000 <= Node.val <= 10000
* Node.random is null or is pointing to some node in the linked list.
*
*
*/
// @lc code=start
/*
// Definition for a Node.
class Node {
public:
int val;
Node* next;
Node* random;
Node(int _val) {
val = _val;
next = NULL;
random = NULL;
}
};
*/
class Solution {
public:
Node* copyRandomList(Node* head) {
if (!head) return nullptr;
unordered_map<Node*, Node*> mymap;
Node dummy(0);
Node* node = &dummy;
Node* pre = &dummy;
node = node->next;
while (head) {
if (mymap.count(head)) {
node = mymap[head];
} else {
node = new Node(head->val);
mymap[head] = node;
}
if (head->random) {
if (mymap.count(head->random)) {
node->random = mymap[head->random];
} else {
Node* newRandom = new Node(head->random->val);
mymap[head->random] = newRandom;
node->random = newRandom;
}
}
pre->next = node;
pre = node;
node = node->next;
head = head->next;
}
return dummy.next;
}
};
// @lc code=end