leetcode 21. Merge Two Sorted Lists
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1.题目
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.将两个有序链表合并成一个有序链表。
2.思路
第一种解法,设置两个指针p1,p2分别指向两个链表的头结点。比较这两个节点值的大小,
如果p1所指向的节点值更小,p1指针所指节点加入目标链表,p1指针后移;
同理,如果p2所指向的节点值更小,p2指针所指节点加入目标链表,p2指针后移;
注意如果有一个链表遍历完之后,把另一个链表剩余部分直接链接在目标链表之后就可以了。
class Solution {public: ListNode* mergeTwoLists(ListNode* a, ListNode* b) { ListNode* p1 = a; ListNode* p2 = b; ListNode* newhead = new ListNode(-1); ListNode* c = newhead; while(p1&&p2){ if(p1->val < p2->val){ c->next = p1; c = c->next; p1=p1->next; } else{ c->next = p2; c=c->next; p2 = p2->next; } } while(p1){c->next = p1;p1 = p1->next;c = c->next;} while(p2){c->next = p2;p2 = p2->next;c = c->next;} return newhead->next; }};
class Solution {public: ListNode* mergeTwoLists(ListNode* a, ListNode* b) { //如果a链表为空,或者当前a节点的值>b节点的值,交换两个节点。这样始终保证a节点的值是当前最小的。因为 //较大的值被交换到b节点去,用于之后的比较 if (!a || b && a->val > b->val) swap(a, b); if (a) a->next = mergeTwoLists(a->next, b); return a; }};
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