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)
pop() // return 1
push(2)
push(3)
min() // return 2
push(1)
min() // return 1LeetCode上的原题，请参见我之前的博客Min Stack.
解法一：
class MinStack { public: MinStack() {} void push(int number) { m1.push(number); if (m2.empty() || m2.top() >= number) { m2.push(number); } } int pop() { if (m1.empty()) return -1; int t = m1.top(); m1.pop(); if (!m2.empty() && m2.top() == t) m2.pop(); return t; } int min() { if (!m2.empty()) return m2.top(); return -1; } private: stack<int> m1, m2; };
解法二：
class MinStack { public: MinStack():mn(INT_MAX) {} void push(int number) { if (number <= mn) { s.push(mn); mn = number; } s.push(number); } int pop() { int t = s.top(); s.pop(); if (t == mn) { mn = s.top(); s.pop(); } return t; } int min() { return mn; } private: int mn; stack<int> s; };
本文转自博客园Grandyang的博客，原文链接：最小栈[LintCode] Min Stack ，如需转载请自行联系原博主。
