155. Min Stack

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.

一刷
题解:在stack中存入的不是x, 而是x-min
例如3,4,2, 2
存入0,1(min=3), -1(min=2), 0
如果当前pop出来的小于0,则恢复原先的min

public class MinStack {
    Stack stack;
    long min = 0;

    /** initialize your data structure here. */
    public MinStack() {
        stack = new Stack<>();
    }
    
    public void push(int x) {
        if(stack.isEmpty()){
            stack.push(0L);
            min = x;
        }
        else{
            stack.push(x-min);
            if(x=0) return (int) (min+peek);
        else{
            long min_ori = min - peek;
            return (int) (peek + min_ori);
        }
    }
    
    public int getMin() {
        return (int)min;
    }
}

/**
 * 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();
 */

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