[Leetcode] Copy List with Random Pointer

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

最開始的想法就是暴力復制，時間復雜度為O(n^2)，寫的時候就感覺要出現事，果不其然，超時了，后來網上看到一個O(n)的算法，非常巧妙的利用了原來鏈表的信息：

該算法更為巧妙，不用保存原始鏈表的映射關系，構建新節點時，指針做如下變化，即把新節點插入到相應的舊節點后面：

 

 

同理分兩步

 

1、構建新節點random指針：new1->random = old1->random->next, new2-random = NULL, new3-random = NULL, new4->random = old4->random->next

 

2、恢復原始鏈表以及構建新鏈表：例如old1->next = old1->next->next,  new1->next = new1->next->next

 

該算法時間復雜度O(N)，空間復雜度O(1)

 

/**
 * Definition for singly-linked list with a random pointer.
 * struct RandomListNode {
 *     int label;
 *     RandomListNode *next, *random;
 *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
 * };
 */
class Solution {
public:
    RandomListNode *copyRandomList(RandomListNode *head) {
        if (head == NULL) return NULL;
        RandomListNode *pos1 = head, *pos2 = head->next;
        while (pos1 != NULL) {
            pos1->next = new RandomListNode(pos1->label);
            pos1->next->next = pos2;
            pos1 = pos2;
            if (pos2 != NULL)
                pos2 = pos2->next;
        }
        pos1 = head;  pos2 = head->next;
        while (pos1 != NULL) {
            if (pos1->random == NULL) {
                pos2->random = NULL;
            } else {
                pos2->random = pos1->random->next;
            }
            pos1 = pos1->next->next;
            if (pos2->next != NULL)
                pos2 = pos2->next->next;
        }
        RandomListNode *res = head->next;
        pos1 = head; pos2 = head->next;
        while(pos2->next != NULL) {
            pos1->next = pos2->next;
            pos1 = pos2;
            if (pos2->next != NULL)
                pos2 = pos2->next;
        }
        pos1->next = NULL;
        pos2->next = NULL;
        return res;
    }
};

 

下面是最開始的超時的暴力復制代碼：

/**
 * Definition for singly-linked list with a random pointer.
 * struct RandomListNode {
 *     int label;
 *     RandomListNode *next, *random;
 *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
 * };
 */
class Solution {
public:
    RandomListNode *copyRandomList(RandomListNode *head) {
        if (head == NULL) return NULL;
        RandomListNode *res = new RandomListNode(head->label);
        RandomListNode *pos1 = head->next, *pos2 = res;
        while (pos1 != NULL) {
            pos2->next = new RandomListNode(pos1->label);
            pos1 = pos1->next;
            pos2 = pos2->next;
        }
        pos1 = head;  pos2 = res;
        RandomListNode *idx1 = head, *idx2 = res;
        while (pos1 != NULL) {
            if (pos1->random == NULL) {
                pos2->random = NULL;
            } else {
                idx1 = head; idx2 = res;
                while (pos1->random != idx1) {
                    idx1 = idx1->next;
                    idx2 = idx2->next;
                }
                pos2->random = idx2;
            }
            pos1 = pos1->next;
            pos2 = pos2->next;
        }
        return res;
    }
};