[传送门]http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4016 [题目大意]初始有n个空栈,现在有如下三种操作: (1) 1 s v 即 s.push(v) (2) 2 s 即 s.pop() 输出弹出的元素,如果栈s为空则输出 "EMPTY" (3) 3 s t 把t栈元素全部移到s栈中,使s的尾部与t的首部相连. 现在有若干上述三种类型的操作,遇到操作2则输出相应内容. [题解]由于站的数量n和操作次数…
Mergeable Stack Time Limit: 2 Seconds Memory Limit: 65536 KB Given initially empty stacks, there are three types of operations: 1 s v: Push the value onto the top of the -th stack. 2 s: Pop the topmost value out of the -th stack, and print that…
Mergeable Stack Time Limit: 2 Seconds Memory Limit: 65536 KB Given initially empty stacks, there are three types of operations: 1 s v: Push the value onto the top of the -th stack. 2 s: Pop the topmost value out of the -th stack, and print tha…
Mergeable Stack Time Limit: 2 Seconds Memory Limit: 65536 KB Given initially empty stacks, there are three types of operations: 1 s v: Push the value onto the top of the -th stack. 2 s: Pop the topmost value out of the -th stack, and print tha…
Implement a stack with min() function, which will return the smallest number in the stack. It should support push, pop and min operation all in O(1) cost. Notice min operation will never be called if there is no number in the stack. Example push(1)…
Implement a stack. You can use any data structure inside a stack except stack itself to implement it. Have you met this question in a real interview? Example push(1) pop() push(2) top() // return 2 pop() isEmpty() // return true push(3) isEmpty()…