惯性聚合 高效追踪和阅读你感兴趣的博客、新闻、科技资讯
阅读原文 在惯性聚合中打开

推荐订阅源

小众软件
小众软件
V
Visual Studio Blog
博客园 - 三生石上(FineUI控件)
Last Week in AI
Last Week in AI
Blog — PlanetScale
Blog — PlanetScale
爱范儿
爱范儿
J
Java Code Geeks
A
About on SuperTechFans
F
Fortinet All Blogs
B
Blog
aimingoo的专栏
aimingoo的专栏
H
Hackread – Cybersecurity News, Data Breaches, AI and More
Engineering at Meta
Engineering at Meta
Y
Y Combinator Blog
有赞技术团队
有赞技术团队
G
Google Developers Blog
Apple Machine Learning Research
Apple Machine Learning Research
V
V2EX
博客园_首页
博客园 - 叶小钗
罗磊的独立博客
OSCHINA 社区最新新闻
OSCHINA 社区最新新闻
D
Docker
云风的 BLOG
云风的 BLOG

博客园 - HonestMan

面试百问 o,1的感悟 公司内部推荐 debain oracle insert method a linked list, find the node that the last node point to. Get balance noe Memory - HonestMan - 博客园 新手开始学习linux print all Permutation of a string An funy question! Google, hire me How to interview a programmer? Binary search tree convert to double linked list. Search in Binary tree spilt a list wirte a function for counting a linked list length Remove repeat char from a string
ShuffleMerge---microsoft's interview question
HonestMan · 2007-09-23 · via 博客园 - HonestMan

Given two lists, merge their nodes together to make one list, taking nodes alternately
between the two lists. So ShuffleMerge() with {1, 2, 3} and {7, 13, 1} should yield {1, 7,
2, 13, 3, 1}. If either list runs out, all the nodes should be taken from the other list. The
solution depends on being able to move nodes to the end of a list as discussed in the
Section 1 review. You may want to use MoveNode() as a helper. Overall, many
techniques are possible: dummy node, local reference, or recursion. Using this function
and FrontBackSplit(), you could simulate the shuffling of cards.
/*
Merge the nodes of the two lists into a single list taking a node
alternately from each list, and return the new list.
*/
struct node* ShuffleMerge(struct node* a, struct node* b) {
// Your code

solution:
struct node* ShuffleMerge(struct node* a, struct node* b) {
    struct node dummy;
    struct node* tail = &dummy;
    dummy.next = NULL;
    while (1) {
        if (a==NULL) { // empty list cases
            tail->next = b;
            break;
        }
        else if (b==NULL) {
            tail->next = a;
            break;
        }
        else { // common case: move two nodes to tail
            tail->next = a;
            tail = a;
            a = a->next;
            tail->next = b;
            tail = b;
            b = b->next;
        }
    }
    return(dummy.next);
}