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子曾经曰过 | hanlijun.net Reviews
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韩利军的随笔
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子曾经曰过,主页 blog,文章归档 archives,关于我 about me,链接 links,学习吧,骚年,weibo,github,用两个栈 stack 实现一个队列 queue,用c 描述,题目大致是这样的,class stack,public,class queue,int dequeue ;,private,stack s1;,stack s2;,况且,由于时间关系,我一般也不要求面试者写代码,只描述清楚思路即可,出这道题,主要考察3点,对于某些具体细节,能不能考虑到 是否足够细致
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子曾经曰过 | hanlijun.net Reviews

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韩利军的随笔

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Blog Archives - 子曾经曰过

http://www.hanlijun.net/blog/archives

A blog for tech and fun. Design credit: Shashank Mehta.

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about-me - 子曾经曰过

http://www.hanlijun.net/aboutme

A blog for tech and fun. E-mail : hancky@sina.com. Github : https:/ github.com/hancky. Design credit: Shashank Mehta.

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memcpy memmove区别和实现 - 子曾经曰过

http://www.hanlijun.net/blog/2014/05/24/memcpy-memmovequ-bie-he-shi-xian

A blog for tech and fun. Void* memcpy(void* dest, const void* src, size t n). Char* d = (char*) dest;. Const char* s = (const char*) src;. Void* memmove(void* dest, const void* src, size t n). Char* d = (char*) dest;. Const char* s = (const char*) src;. Start at beginning of s. Else if (s d). Start at end of s. 内存低端 ——-s——– ——-d——– 内存高端 start at end of s. 内存低端 ——-s— = —d——– 内存高端 start at end of s. 内存低端 ——-sd——– 内存高端 do nothing. 转自 http:/ www.cnblogs.com/kekec/archive/2011/07/22/2114107.html.

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links - 子曾经曰过

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子曾经曰过

A blog for tech and fun. Void Push(int x); / Push an element in stack;. Int Pop(); / Pop an element out of stack;. Int Count() const; / Return the number of the elements in stack;. Void Enqueue(int x);. 这道题应该不算难,比起 编程之美 中微软那些什么 翻烙饼 的面试题,难度上差远了。 出队时,将s1的元素逐个 倒入 弹出并压入 s2,将s2的顶元素弹出作为出队元素,之后再将s2剩下的元素逐个 倒回 s1。 即 在出队时,将s1的元素逐个 倒入 s2时,原在s1栈底的元素,不用 倒入 s2 即只 倒 s1.Count()-1个 ,可直接弹出作为出队元素返回。 上述思路,有些变种,如 入队时,先判断s1是否为空,如不为空,说明所有元素都在s1,此时将入队元素直接压入s1 如为空,要将s2的元素逐个 倒回 s1,再压入入队元素。 这个思路,避免了反复 倒 栈,仅在需要时才 倒 一次。 Else if (s d).

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