LeetCode(21)-Merge Two Sorted Lists

问题描述：

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.

问题分析：

合并两个有序的链表，考察对链表的操作，比较经典的题目

解题方法；

/** * 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) {        ListNode *start, *cur;        if(l1 == NULL)            return l2;        if(l2 == NULL)            return l1;                if(l1->val < l2->val ){            cur = start = l1;            l1 = l1->next;        }        else{            cur = start = l2;            l2 = l2->next;        }        while(l1 != NULL && l2 != NULL){            if(l1->val < l2->val ){                cur->next = l1;                cur = l1;                l1 = l1->next;            }            else{                cur->next = l2;                cur = l2;                l2 = l2->next;            }        }        if(l1 !=NULL)            cur->next = l1;        else            cur->next = l2;        return start;    }};

一开始不懂为什么要定义两个指针，start和cur，后来尝试返回cur时才明白，start一直指向合并链表的头，cur一直在变，所以不能用cur代替。

/** * 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) {        ListNode* head = new ListNode(0); //开始我让head = NULL,总是编译不通过，后来改成这样通过了        ListNode* prev = head;                if(l1 == NULL) {            return l2;        }        if(l2 == NULL) {           return l1;         }                 //prev = head;                while(l1 != NULL && l2 != NULL){            if(l1->val < l2->val){                prev->next = l1;                l1 = l1->next;                prev = prev->next;            }            else{                prev->next = l2;                l2 = l2->next;                prev = prev->next;            }        }                if(l1 != NULL){             prev->next = l1;        }        if(l2 != NULL) {            prev->next = l2;         }                  return head->next;    }};

这是我第二次做，下次要注意定义链表头节点时要初始化，不能等于NULL.