155. Min Stack - Easy

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.

 

M1: 用两个stack，s1 push 当前元素，s2同步 push 当前最小值，pop也同步

push(), pop(), top(), getMin()　　-- time: O(1), space: O(n)

class MinStack {
    LinkedList<Integer> stack;
    LinkedList<Integer> minStack;

    /** initialize your data structure here. */
    public MinStack() {
        stack = new LinkedList<>();
        minStack = new LinkedList<>();
    }
    
    public void push(int x) {
        stack.push(x);
        if(minStack.isEmpty()) {
            minStack.push(x);
        } else {
            minStack.push(Math.min(x, minStack.peek()));
        }
    }
    
    public void pop() {
        stack.pop();
        minStack.pop();
    }
    
    public int top() {
        return stack.peek();
    }
    
    public int getMin() {
        return minStack.peek();
    }
}

/**
 * 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();
 */

 

M2: 用一个stack，常数global min存最小值

push: 如果当前元素会改变global min，先push一个旧的global min, 更新global min，再把当前元素入栈

pop: 如果pop出的元素 = global min，再pop一次，global min = 第二次pop的数，即旧的global min

class MinStack {
    LinkedList<Integer> s;
    int globalMin;

    /** initialize your data structure here. */
    public MinStack() {
        s = new LinkedList<>();
        globalMin = Integer.MAX_VALUE;
    }
    
    public void push(int x) {
        if(x <= globalMin) {
            s.push(globalMin);
            globalMin = x;
        }
        s.push(x);
    }
    
    public void pop() {
        if(s.pop() == globalMin) {
            globalMin = s.pop();
        }
    }
    
    public int top() {
        return s.peek();
    }
    
    public int getMin() {
        return globalMin;
    }
}

/**
 * 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();
 */

 

M3: 用两个stack，有大量重复元素的时候，s2可以存这样的pair <val, s1.size()> ，节约空间

class MinStack {
    LinkedList<Integer> stack;
    LinkedList<int[]> minStack;

    /** initialize your data structure here. */
    public MinStack() {
        stack = new LinkedList<>();
        minStack = new LinkedList<>();
    }
    
    public void push(int x) {
        stack.push(x);
        if(minStack.isEmpty()) {
            minStack.push(new int[] {x, stack.size()});
        } else {
            if(x < minStack.peek()[0]) {
                minStack.push(new int[] {x, stack.size()});
            }
        }
    }
    
    public void pop() {
        if(stack.size() == minStack.peek()[1]) {
            stack.pop();
            minStack.pop();
        } else {
            stack.pop();
        }
    }
    
    public int top() {
        return stack.peek();
    }
    
    public int getMin() {
        return minStack.peek()[0];
    }
}

/**
 * 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();
 */