1.描述
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.
Example:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
2.分析
用两个栈，一个用来正常存储，另一个用来存储当前状态下的最小值。
3.代码
class MinStack {
private:
stack<int> s1,s2;
public:
/** initialize your data structure here. */
MinStack() {
}
void push(int x) {
s1.push(x);
if (s2.empty()) {
s2.push(x);
} else {
s2.push(s2.top() < x ? s2.top() : x);
}
}
void pop() {
if (s1.empty() || s2.empty()) exit(-1);
s1.pop();
s2.pop();
}
int top() {
if (s1.empty() || s2.empty()) exit(-1);
return s1.top();
}
int getMin() {
if (s1.empty() || s2.empty()) exit(-1);
return s2.top();
}
};
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
