LeetCode-Min Stack

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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.

所以就需要节省空间，不是每次push的时候都push最小值，而是每次最小值发生改变的时候才push。代码如下：

class MinStack {
    LinkedList<Integer> stack = new LinkedList<Integer>();
    LinkedList<Integer> minStack = new LinkedList<Integer>();
    public void push(int x) {
        if (x <= getMin()) {
            minStack.push(x);
        }
        stack.push(x);
    }

    public void pop() {
        if (stack.pop() == getMin()) {
            minStack.pop();
        } 
    }

    public int top() {
        return stack.peek();
    }

    public int getMin() {
        if (minStack.isEmpty()) {
            return Integer.MAX_VALUE;
        } else {
            return minStack.peek();
        }
    }
}

class MinStack {
    private long min = 0;
    private LinkedList<Long> stack = new LinkedList<>();
    public void push(int x) {
        if (stack.isEmpty()) {
            min = x;
            stack.push(0L);
        } else {
            stack.push(x-min);
            if (x-min < 0) {
                min = x;
            }
        }
    }

    public void pop() {
       if (stack.peek() < 0) {
           min -= stack.peek();
       }
       stack.pop();
    }

    public int top() {
        if (stack.peek() < 0) {
            return (int) min;
        } 
        return (int) (min+stack.peek());
    }

    public int getMin() {
       return (int) min;
    }
}

每次并不是push原数据，而是与min坐过差值之后在存放，good idea！！！

LeetCode-Min Stack

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原文地址：http://blog.csdn.net/my_jobs/article/details/47612395