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leetcode--155. Min Stack

emp sta n) ports long public after min() minimum

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.


將min設成long是因為可能是Integer.MAX_VALUE-Integer.MIN_VALUE
棧裏存儲的是push的值與最小值的差值。如果,差值小於零,說明push的值比最小值還要小,當取出的時候如果是小於零的值,取出的值就是min同時更新min
class MinStack {
    long min;
    Stack<Long> st;
    /** initialize your data structure here. */
    public MinStack() {
        st=new Stack<>();
    }
    
    public void push(int x) {
        if(st.isEmpty()){
            st.push(0L);
            min=x;
        }
        else{
            st.push(x
-min); if(x<min) min=x; } } public void pop() { if(st.isEmpty())return; long p=st.pop(); if(p<0) min=min-p; } public int top() { long t=st.peek(); if(t<0) return (int)min;
else return (int)(t+min); } public int getMin() { return (int)min; } }

這還有個絕妙的做法

class MinStack {
    int min = Integer.MAX_VALUE;
    Stack<Integer> stack = new Stack<Integer>();
    public void push(int x) {
        // only push the old minimum value when the current 
        // minimum value changes after pushing the new value x
        if(x <= min){          
            stack.push(min);
            min=x;
        }
        stack.push(x);
    }

    public void pop() {
        // if pop operation could result in the changing of the current minimum value, 
        // pop twice and change the current minimum value to the last minimum value.
        if(stack.pop() == min) min=stack.pop();
    }

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

    public int getMin() {
        return min;
    }
}

leetcode--155. Min Stack