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中所含最小元素的min函数（时间复杂度应为O（1））。

 1 class MinStack:
 2     def __init__(self):
 3         self.stack = []
 4         self.min_stack = []
 5 
 6     def push(self, x: int) -> None:
 7         self.stack.append(x)
 8         if len(self.min_stack) == 0 or x <= self.min_stack[-1]:
 9             self.min_stack.append(x)
10 
11     def pop(self) -> None:
12         x = self.stack.pop()
13         if x == self.min_stack[-1]:
14             self.min_stack.pop()
15 
16     def top(self) -> int:
17         return self.stack[-1]
18 
19     def getMin(self) -> int:
20         return self.min_stack[-1]

Java版代码，leetcode地址：

 1 class MinStack {
 2 
 3         Stack<Integer> normalStack;
 4         Stack<Integer> minStack;
 5 
 6         /** initialize your data structure here. */
 7         public MinStack() {
 8             normalStack = new Stack<Integer>();
 9             minStack = new Stack<Integer>();
10         }
11 
12         public void push(int x) {
13             normalStack.add(x);
14             if (minStack.isEmpty() || minStack.peek() >= x) {
15                 minStack.add(x);
16             }
17         }
18 
19         public void pop() {
20             int x = normalStack.pop();
21             if (x == minStack.peek()) {
22                 minStack.pop();
23             }
24         }
25 
26         public int top() {
27             return normalStack.peek();
28         }
29 
30         public int min() {
31             return minStack.peek();
32         }
33     }

posted on 2019-06-13 06:02 Sempron2800+ 阅读(132) 评论(0) 编辑收藏举报