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.
min operation will never be called if there is no number in the stack.
push(1)
pop() // return 1
push(2)
push(3)
min() // return 2
push(1)
min() // return 1
LeetCode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; };
原文：http://www.cnblogs.com/grandyang/p/6060577.html