剑指 Offer 30. 包含min函数的栈
定义栈的数据结构,请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中,调用 min、push 及 pop 的时间复杂度都是 O(1)。
链接:https://leetcode-cn.com/problems/bao-han-minhan-shu-de-zhan-lcof/
示例:
MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.min(); --> 返回 -3. minStack.pop(); minStack.top(); --> 返回 0. minStack.min(); --> 返回 -2.
提示:
各函数的调用总次数不超过 20000 次
解法1 一个栈
min()方法的效率比较低,就是挨个比较。
class MinStack { Stack<Integer> stack = null; /** initialize your data structure here. */ public MinStack() { stack = new Stack<>(); } public void push(int x) { stack.push(x); } public void pop() { stack.pop(); } public int top() { return stack.peek(); } public int min() { int min = Integer.MAX_VALUE; for (Integer integer : stack) { if(min > integer) min = integer; } return 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.min(); */
解法2 双栈
class MinStack { Stack<Integer> stack = null; Stack<Integer> stackB = null; /** initialize your data structure here. */ public MinStack() { stack = new Stack<>(); stackB = new Stack<>(); } public void push(int x) { stack.push(x); if(stackB.isEmpty() || stackB.peek() >= x) stackB.push(x); } public void pop() { if(stack.pop().equals(stackB.peek())) stackB.pop(); } public int top() { return stack.peek(); } public int min() { return stackB.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.min(); */