Merge Two Sorted Lists
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题目描述
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
思路
要把两个已经排序的list(L1, L2)合并并且排好序。那么首先比较的就是两个队列的head,然后将value较小的head作为新list的head。
这里我们假设是L1->head成为了newlist->head,所以newlist->val=L1->value,然后L1=L1->next,但是L2不变。继续比较L1和L2的value值大小。
这里假设L2->val< L1->val,那么newlist->val=L2->val,然后L2=L2->next,L1不变。
重复操作知道L1,L2遍历完成。
class Solution {public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if (l1 == NULL) return l2; if (l2 == NULL) return l1; ListNode* head = NULL; if (l1->val < l2->val) { head = l1; l1 = l1->next; } else { head = l2; l2 = l2->next; } ListNode* p = head; while (l1 && l2) { if (l1->val < l2->val) { p->next = l1; l1 = l1->next; } else { p->next = l2; l2 = l2->next; } p = p->next; } if(l1) p->next = l1; else p->next = l2; return head; }};
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