LC.155. Min Stack(非优化,两个stack 同步 + -)
https://leetcode.com/problems/min-stack/description/
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.
corner cases:
1)If the stack is empty, min() should return -1.
2)pop() - remove and return the top element, if the stack is empty, return -1
3)push(int element) - push the element to top
top() - return the top element without remove it, if the stack is empty, return -1
4) min() - return the current min value in the stack.
这种写法 stack 1 stack 2 同步操作,stack 1 进来一个 stack2 也要进来一个。如果有特别多重复的数进来的情况,并且是一整块一样的数进来,那是可以被优化的。
1111111 222222 -1-1-1-1 111111 -2-2-2-2-2 33333
如果 进来的数字是交叉的,则优化空间有限: 1212121212
1 public class LC_155_MinStack { 2 3 private Deque<Integer> stack ; 4 private Deque<Integer> minStack ; 5 6 /** initialize your data structure here. */ 7 public LC_155_MinStack() { 8 stack = new LinkedList<>() ; 9 minStack = new LinkedList<>(); 10 } 11 /* 12 stack -2 0 -3 13 min -2 -2 -3 14 * */ 15 public void push(int x) { 16 stack.push(x); 17 if (minStack.isEmpty()){ 18 minStack.push(x); 19 } else{ 20 if (minStack.peek() < x){ 21 minStack.push(minStack.peek()); 22 } else{ 23 minStack.push(x); 24 } 25 } 26 } 27 28 public void pop() { 29 if (stack.isEmpty()) return ; 30 stack.pop(); 31 minStack.pop(); 32 } 33 //peek 34 public int top() { 35 if (stack.isEmpty()) return -1 ; 36 return stack.peek() ; 37 } 38 39 public int getMin() { 40 if (minStack.isEmpty()) return -1 ; 41 return minStack.peek(); 42 } 43 }