[剑指offer] 合并两个排序的链表

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

---

思路比较简单，主要是对指针的操作要考虑完全。

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        if (!pHead1 && !pHead2) return NULL;
        if (!pHead1) return pHead2;
        if (!pHead2) return pHead1;
        ListNode* re, *result;    // re不断更新来向尾部创建节点，result存储头指针
        if (pHead1->val > pHead2->val) {
            re = new ListNode(pHead2->val);
            pHead2 = pHead2->next;
        } else {
            re = new ListNode(pHead1->val);
            pHead1 = pHead1->next;
        }
        result = re;
        while (pHead1 || pHead2) {
            if (pHead1 && pHead2) {
                if ((pHead1->val) > (pHead2->val)) {
                    re->next = new ListNode(pHead2->val);
                    pHead2 = pHead2->next;
                } else {
                    re->next = new ListNode(pHead1->val);
                    pHead1 = pHead1->next;
                }
                re = re->next;
            } else if (pHead1) {
                while (pHead1) {
                    re->next = new ListNode(pHead1->val);
                    pHead1 = pHead1->next;
                    if (pHead1) re = re->next;
                }
            } else {
                while (pHead2) {
                    re->next = new ListNode(pHead2->val);
                    pHead2 = pHead2->next;
                    if (pHead2) re = re->next;
                }
            }
        }
        return result;
    }
};