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HANLILILARIN BULUSMA NOKTASINA HOSGELDINIZ | hanlikoyu.com Reviews
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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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