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Method 1: Dual stack simulation minimum stack: time O(1), space O(n)
answer:
- To ensure that the time complexity of push(), pop(), min() and other functions are all O(1), so we manage a minimum stack dedicated to storing the smallest value in the current stack
- push(): Push directly into the stack, min stack: Determine the size of the top element and the inserted element, and insert if the inserted element is smaller or equal
- pop(): directly delete the push stack, min stack: determine whether the top element of the stack is the same as the element deleted by the push stack, delete if the same
class MinStack {
private:
stack<int> push_stc;
stack<int> min_stc;
public:
/** initialize your data structure here. */
MinStack() {
}
void push(int x) {
push_stc.push(x);
if (min_stc.empty() || min_stc.top() >= x)
{
min_stc.push(x);
}
}
void pop() {
if (push_stc.empty())
return;
if (push_stc.top() == min_stc.top())
min_stc.pop();
push_stc.pop();
}
int top() {
if (push_stc.empty())
return -1;
return push_stc.top();
}
int min() {
if (min_stc.empty())
return -1;
return min_stc.top();
}
};