Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
- push(x) -- Push element x onto stack.
- pop() -- Removes the element on top of the stack.
- top() -- Get the top element.
- getMin() -- Retrieve the minimum element in the stack.
Subscribe to see which companies asked this question
解法:还是用两个栈s1和s2来模拟。s1按顺序保存数据,而s2保存当前最小值。在进栈时,如果s2为空或者当前进栈值不大于s2栈顶值,则也进s2;出栈时,如果当前出栈值恰好为s2的栈顶值,则s2也要出栈。
class MinStack { public: void push(int x) { s1.push(x); if (s2.empty() || x <= s2.top()) s2.push(x); } void pop() { if (!s1.empty()) { if (!s2.empty() && s1.top() == s2.top()) s2.pop(); s1.pop(); } } int top() { if (!s1.empty()) return s1.top(); return 0; } int getMin() { if (!s2.empty()) return s2.top(); return 0; } private: std::stack<int> s1, s2; };