我的输入输出
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
是
[null,null,null,null,-3,null,-3,-2]
预期是
[null,null,null,null,-3,null,0,-2]
我期望"top"
方法是我的问题所在,但是我无法弄清楚自己在做什么错。有人可以为我指出吗?我正在使用两个堆栈来解决问题。通过将其值与第一个堆栈的值进行比较,将一个堆栈压入输入,并将一个minStack压入输入。
/**
* initialize your data structure here.
*/
let MinStack = function() {
this.stack = new Stack();
this.minStack = new Stack();
};
/**
* @param {number} x
* @return {void}
*/
MinStack.prototype.push = function(x) {
this.stack.push(x);
if (this.minStack.size === 0) {
this.minStack.push(x);
} else if (x <= this.minStack.peek()) {
this.minStack.push(x);
}
};
/**
* @return {void}
*/
MinStack.prototype.pop = function() {
let popped = this.stack.peek();
if (popped === this.minStack.peek()) {
this.minStack.pop()
}
};
/**
* @return {number}
*/
MinStack.prototype.top = function() {
return this.stack.peek();
};
/**
* @return {number}
*/
MinStack.prototype.getMin = function() {
return this.minStack.peek();
};
class Stack {
constructor() {
this.storage = {};
this.size = 0;
}
push(val) {
this.storage[this.size] = val;
this.size++;
}
pop() {
let top = this.storage[this.size - 1];
delete this.storage[this.size - 1];
this.size--;
return top;
}
peek() {
return this.storage[this.size - 1];
}
getSize() {
return this.size;
}
empty() {
return this.size === 0;
}
}
答案 0 :(得分:0)
好问题!
不确定您的错误,我们可以使用JavaScript数组解决此问题。这样就可以了:
var MinStack = function() {
this.stack = [];
};
MinStack.prototype.push = function(x) {
this.stack.push({
value: x,
min: this.stack.length === 0 ? x : Math.min(x, this.getMin()),
});
};
MinStack.prototype.pop = function() {
this.stack.pop();
};
MinStack.prototype.top = function() {
return this.stack[this.stack.length - 1].value;
};
MinStack.prototype.getMin = function() {
return this.stack[this.stack.length - 1].min;
};
如果您想使用自己的Stack()
,请将其插入上面的代码中:
this.stack = [];
,它应该可以正常工作。为此,您将不得不编写一个min()
方法,我想它会像这样:
class MyOwnMinStack {
constructor() {
this.storage = {};
this.size = 0;
this.currMin = null
}
push(val) {
this.storage[this.size] = val;
this.size++;
}
pop() {
let top = this.storage[this.size - 1];
delete this.storage[this.size - 1];
this.size--;
return top;
}
peek() {
return this.storage[this.size - 1];
}
getSize() {
return this.size;
}
empty() {
return this.size === 0;
}
min() {
// To do
return this.currMin;
}
}
每次我们push()
或pop()
时,都会跟踪this.currMin
,以便一旦{{ 1}}。
在这里,我将复制其他方法来解决其他语言的问题,可能会帮助您解决问题。例如,此c++方法使用两个堆栈:
O(1)
min()
class MinStack {
private:
stack<int> stack_one;
stack<int> stack_two;
public:
void push(int x) {
stack_one.push(x);
if (stack_two.empty() || x <= getMin())
stack_two.push(x);
}
void pop() {
if (stack_one.top() == getMin())
stack_two.pop();
stack_one.pop();
}
int top() {
return stack_one.top();
}
int getMin() {
return stack_two.top();
}
};