Description:
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.
My code:
/**
* initialize your data structure here.
*/
var MinStack = function() {
this.dataStore = [];
this.length = 0;
};
/**
* @param {number} x
* @return {void}
*/
MinStack.prototype.push = function(x) {
this.dataStore[this.length++] = x;
};
/**
* @return {void}
*/
MinStack.prototype.pop = function() {
let temp = this.dataStore[this.length];
this.length--;
this.dataStore.length--;
return temp;
};
/**
* @return {number}
*/
MinStack.prototype.top = function() {
return this.dataStore[this.length - 1];
};
/**
* @return {number}
*/
MinStack.prototype.getMin = function() {
let min = Number.MAX_VALUE;
for(let i = 0; i <= this.length; i++) {
if(min > this.dataStore[i]) {
min = this.dataStore[i];
}
}
return 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()
*/
Note: 雖然是寫出來了，但是Submissions那里看到run time是1000多…… 還需要優(yōu)化溃列，getMin可以舍棄for循環(huán)的
