[LeetCode-21] Merge Two Sorted Lists（合并两个有序链表）

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.

【分析】合并两个有序链表到一个链表里

代码：

/** * Definition for singly-linked list. **/ struct ListNode {      int val;      struct ListNode *next; };struct ListNode* mergeTwoLists(struct ListNode* FirstLists,                                     struct ListNode* SecondLists){    /*1.异常处理*/    if(!FirstLists && !SecondLists)        return NULL;    else if(!FirstLists && SecondLists)        return SecondLists;    else if(FirstLists && !SecondLists)        return FirstLists;    struct ListNode *FirstListsTemp = FirstLists;    struct ListNode *SecondListsTemp = SecondLists;    struct ListNode *head = NULL;    struct ListNode *curNode = NULL;    /*2.合并两个有序的链表*/    while(FirstListsTemp&&SecondListsTemp) {        /*3.获得头结点，作为返回值*/        if(head == NULL) {              if(FirstListsTemp->val >= SecondListsTemp->val) {                curNode = SecondListsTemp;                head = curNode;                SecondListsTemp  = SecondListsTemp->next;                curNode->next = NULL;            }            else{                curNode = FirstListsTemp;                head = curNode;                FirstListsTemp = FirstListsTemp->next;                curNode->next = NULL;            }        }        else {            if(FirstListsTemp->val >= SecondListsTemp->val) {                curNode->next = SecondListsTemp;                curNode = curNode->next;                SecondListsTemp = SecondListsTemp->next;                curNode->next = NULL;            }            else{                curNode->next = FirstListsTemp;                curNode = curNode->next;                FirstListsTemp = FirstListsTemp->next;                curNode->next = NULL;            }        }    }    /*4.剩下的有序链表*/    if(FirstListsTemp) {        curNode->next = FirstListsTemp;    }    if(SecondListsTemp) {        curNode->next = SecondListsTemp;    }    return head;}