【LeetCode】155. Min Stack

题目：

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.

题解：

使用两个栈，一个数据栈，一个用来存储每一步的最小值

Solution 1 

class MinStack {
public:
    MinStack() {
        
    }
    
    void push(int x) {
        s1.push(x);
        if(s2.empty() || s2.top() >= x)
            s2.push(x);
    }
    
    void pop() {
        if(s1.top() == s2.top()){
            s2.pop();
        }
        s1.pop();
    }
    
    int top() {
        return s1.top();
    }
    
    int getMin() {
        return s2.top();
    }
    stack<int> s1, s2;
};

只用一个栈，不过需要额外的整形变量记录最小值。

Solution 2 

class MinStack {
public:
    /** initialize your data structure here. */
    MinStack() {
        
    }
    
    void push(int x) {
        if(x <= min_val){
            s.push(min_val);
            min_val = x;
        }
        s.push(x);
    }
    
    void pop() {
        if(s.top() == min_val){
            s.pop();
            min_val = s.top();
        }
        s.pop();
    }
    
    int top() {
        return s.top();
    }
    
    int getMin() {
        return min_val;
    }
    int min_val = INT_MAX;
    stack<int> s;
};