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本文地址: http://blog.csdn.net/caroline_wendy/article/details/29352997
题目:
合并有序链表, 给定两个升序的链表, 返回一个合并之后的升序链表.
节点结构:
struct Node{
int val;
Node *next;
};
要求实现的函数:
Node* mergeList(Node *list_a, Node* list_b)
代码:
/* * test.cpp * * Created on: 2014.04.24 * Author: Spike */ /*eclipse cdt, gcc 4.8.1*/ #include <iostream> #include <vector> using namespace std; struct Node{ int val; Node *next; }; Node* mergeList(Node *list_a, Node* list_b) { if (list_a == NULL) //递归的终止条件 return list_b; else if (list_b == NULL) return list_a; Node* pMergedHead = NULL; //合并后的链表 if (list_a->val < list_b->val) { pMergedHead = list_a; //指向头结点 pMergedHead->next = mergeList(list_a->next, list_b); //递归 } else { pMergedHead = list_b; pMergedHead->next = mergeList(list_a, list_b->next); } return pMergedHead; } Node* initList(const std::vector<int>& vi) { Node* pHead = new Node; Node* pTemp = pHead; for (std::size_t i=0; i<vi.size(); ++i) { pTemp->val = vi[i]; if (i != vi.size()-1) { //非尾结点 Node* pNode = new Node; pTemp->next = pNode; pTemp = pTemp->next; } } pTemp->next = NULL; return pHead; } void printList(Node* L) { Node* pTemp = L; while (pTemp->next != NULL) { std::cout << pTemp->val << " "; pTemp = pTemp->next; } std::cout << pTemp->val << " "; //打印最后一个值 std::cout << std::endl; } int main(void) { std::vector<int> via = {1, 2, 3, 4, 5, 13}; std::vector<int> vib = {2, 4, 5, 7, 9, 11}; Node* list_a = initList(via); Node* list_b = initList(vib); std::cout << "list_a = "; printList(list_a); std::cout << "list_b = "; printList(list_b); Node* list_merge = mergeList(list_a, list_b); std::cout << "list_merge = "; printList(list_merge); return 0; }
list_a = 1 2 3 4 5 13 list_b = 2 4 5 7 9 11 list_merge = 1 2 2 3 4 4 5 5 7 9 11 13
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原文地址:http://www.cnblogs.com/lcchuguo/p/4557079.html