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)。
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：
注意：本题与主站 155 题相同：https://leetcode.cn/problems/min-stack/
原站题解
golang 解法, 执行用时: 60 ms, 内存消耗: 8.2 MB, 提交时间: 2020-11-09 23:25:14
type MinStack struct {
stack []int
length int
}
/** initialize your data structure here. */
func Constructor() MinStack {
return MinStack{stack: []int{}, length: 0}
}
func (this *MinStack) Push(x int) {
this.length += 1
this.stack = append(this.stack, x)
}
func (this *MinStack) Pop() {
this.length -= 1
this.stack = this.stack[:this.length]
}
func (this *MinStack) Top() int {
return this.stack[this.length-1]
}
func (this *MinStack) Min() int {
min := this.stack[0]
for i := 1; i < this.length; i++ {
if min > this.stack[i] {
min = this.stack[i]
}
}
return min
}
/**
* Your MinStack object will be instantiated and called as such:
* obj := Constructor();
* obj.Push(x);
* obj.Pop();
* param_3 := obj.Top();
* param_4 := obj.Min();
*/
python3 解法, 执行用时: 560 ms, 内存消耗: 16.6 MB, 提交时间: 2020-11-09 23:12:54
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
def push(self, x: int) -> None:
self.stack.append(x)
def pop(self) -> None:
self.stack.pop()
def top(self) -> int:
return self.stack[-1]
def min(self) -> int:
return min(self.stack)
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.min()