LeetCode

题意

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.

解法

两个栈，一个栈存原始数据，一个栈始终保留当前最小元素

实现

class MinStack {private:stack<int> oriStack;stack<int> minStack;public:    /** initialize your data structure here. */    MinStack() {    }    void push(int x) {        oriStack.push(x);        if(minStack.empty()) minStack.push(x);        else if(minStack.top() < x) minStack.push(minStack.top());        else minStack.push(x);    }    void pop() {        if(oriStack.empty()) return;        oriStack.pop();        minStack.pop();    }    int top() {        if(oriStack.empty()) return -1;        return oriStack.top();    }    int getMin() {        if(oriStack.empty()) return -1;        return minStack.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(); */