Lintcode: Min Stack

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.

Have you met this question in a real interview? 

Yes

Example

Operations: push(1), pop(), push(2), push(3), min(), push(1), min() Return: 1, 2, 1

Note

min operation will never be called if there is no number in the stack

Reference my leetcode solution for the same problem:

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class MinStack { private: vector<int> data; int minV; public: MinStack() { // do initialization if necessary } void push(int number) { // write your code here if(data.size() == 0) minV = number; else{ int prev = minV + data.back(); if(prev < minV) minV = prev; } data.push_back(number - minV); } int pop() { // write your code here int res = data.back() + minV; data.pop_back(); if(data.back() < 0) minV = minV - data.back(); return res; } int min() { // write your code here return ((data.back() + minV) < minV)?(data.back() + minV) : minV; } };

Lintcode second:
Space: O(n)

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class MinStack { public: vector<int> data; vector<int> minData; MinStack() { // do initialization if necessary } void push(int number) { // write your code here data.push_back(number); if(minData.empty() || number <= minData.back()) minData.push_back(number); } int pop() { // write your code here if(data.back() == minData.back()) minData.pop_back(); int res = data.back(); data.pop_back(); return res; } int min() { // write your code here return minData.back(); } };