Problem
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.
Solution 1
/**
* initialize your data structure here.
*/
var MinStack = function() {
this.min = [];
this.stack = [];
};
/**
* @param {number} x
* @return {void}
*/
MinStack.prototype.push = function(x) {
if (this.min.length === 0 || x <= this.min[this.min.length - 1]) this.min.push(x);
this.stack.push(x);
};
/**
* @return {void}
*/
MinStack.prototype.pop = function() {
var val = this.stack.pop();
if (val === this.min[this.min.length - 1]) this.min.pop();
};
/**
* @return {number}
*/
MinStack.prototype.top = function() {
return this.stack.length ? this.stack[this.stack.length - 1] : 0;
};
/**
* @return {number}
*/
MinStack.prototype.getMin = function() {
return this.min.length ? this.min[this.min.length - 1] : 0;
};
/**
* Your MinStack object will be instantiated and called as such:
* var obj = Object.create(MinStack).createNew()
* obj.push(x)
* obj.pop()
* var param_3 = obj.top()
* var param_4 = obj.getMin()
*/
Explain:
两个栈，一个存值，一个存最小值。
Complexity:
- Time complexity :
- Space complexity :
Solution 2
/**
* initialize your data structure here.
*/
var MinStack = function() {
this.min = 0;
this.length = 0;
this.stack = [];
};
/**
* @param {number} x
* @return {void}
*/
MinStack.prototype.push = function(x) {
if (!this.length || x <= this.min) {
this.stack.push(this.min);
this.min = x;
}
this.stack.push(x);
this.length++;
};
/**
* @return {void}
*/
MinStack.prototype.pop = function() {
if (!this.length) return;
var val = this.stack.pop();
if (val === this.min) {
this.min = this.stack.pop();
}
this.length--;
};
/**
* @return {number}
*/
MinStack.prototype.top = function() {
return this.length ? this.stack[this.stack.length - 1] : 0;
};
/**
* @return {number}
*/
MinStack.prototype.getMin = function() {
return this.min;
};
/**
* Your MinStack object will be instantiated and called as such:
* var obj = Object.create(MinStack).createNew()
* obj.push(x)
* obj.pop()
* var param_3 = obj.top()
* var param_4 = obj.getMin()
*/
Explain:
一个栈，同时存值和最小值。
Complexity:
- Time complexity :
- Space complexity :
Solution 3
/**
* initialize your data structure here.
*/
var MinStack = function() {
this.Node = function (val) {
this.val = val;
this.min = 0;
this.next = null;
};
this.head = null;
};
/**
* @param {number} x
* @return {void}
*/
MinStack.prototype.push = function(x) {
var node = new this.Node(x);
if (this.head) {
node.next = this.head;
node.min = Math.min(x, this.head.min);
} else {
node.min = x;
}
this.head = node;
};
/**
* @return {void}
*/
MinStack.prototype.pop = function() {
if (this.head) this.head = this.head.next;
};
/**
* @return {number}
*/
MinStack.prototype.top = function() {
return this.head ? this.head.val : 0;
};
/**
* @return {number}
*/
MinStack.prototype.getMin = function() {
return this.head ? this.head.min : 0;
};
/**
* Your MinStack object will be instantiated and called as such:
* var obj = Object.create(MinStack).createNew()
* obj.push(x)
* obj.pop()
* var param_3 = obj.top()
* var param_4 = obj.getMin()
*/
Explain:
用链表模拟栈，每个节点存储值 val，最小值 min，下一个节点 next。
Complexity:
- Time complexity :
- Space complexity :