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LeetCode算法笔记--最小栈

算法题 热题HOT100 栈
字数统计: 244阅读时长: 1 min
2021/02/27 Share

LeetCode算法笔记-Day72

155. Min Stack

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

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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.
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Example:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

最小栈

Snip20201027_1.png

Answer：

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/**
 * initialize your data structure here.
 */
var MinStack = function() {
    this._stack = [];
    this._min_stack = [Infinity];
};

/** 
 * @param {number} x
 * @return {void}
 */
MinStack.prototype.push = function(x) {
    this._stack.push(x);
    this._min_stack.push(Math.min(this._min_stack[this._min_stack.length-1],x)) ;
};

/**
 * @return {void}
 */
MinStack.prototype.pop = function() {
    this._stack.pop();
    this._min_stack.pop();
};

/**
 * @return {number}
 */
MinStack.prototype.top = function() {
    return this._stack[this._stack.length-1]
};

/**
 * @return {number}
 */
MinStack.prototype.getMin = function() {
    return this._min_stack[this._min_stack.length-1]
};

/**
 * Your MinStack object will be instantiated and called as such:
 * var obj = new MinStack()
 * obj.push(x)
 * obj.pop()
 * var param_3 = obj.top()
 * var param_4 = obj.getMin()
 */
CATALOG
  1. 1. LeetCode算法笔记-Day72
    1. 1.1. 155. Min Stack
    2. 1.2. 最小栈
      1. 1.2.1. Answer：
Total : 137
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