LeetCode—Merge Two Sorted Lists融合兩個有序單鏈表
阿新 • • 發佈:2019-01-10
首先兩個單鏈表是有序的
在融合兩個單鏈表的時候,如果想到的是在一個序列上進行增減,那麼會非常麻煩
這裡一定要單獨開一個序列頭進行儲存,不一定需要開闢記憶體,主要是一個概念
其實方法感覺和歸併演算法的merge都是一個概念
提供劍指offer上的遞迴的解決辦法/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { if(l1 == NULL) { return l2; } else 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 *phead = head; while(l1 && l2) { if(l1->val < l2->val) { head->next = l1; head = head->next; l1 = l1->next; } else { head->next = l2; head = head->next; l2 = l2->next; } } if(l1) { head->next = l1; } if(l2) { head->next = l2; } return phead; } };
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { if(l1 == NULL) { return l2; } else if(l2 == NULL) { return l1; } ListNode *head = NULL; if(l1->val < l2->val) { head = l1; head->next = mergeTwoLists(l1->next,l2); } else { head = l2; head->next = mergeTwoLists(l1,l2->next); } return head; } };