【LeetCode热题100（35/100）】LRU 缓存

题目地址： 链接
思路： 利用 this.map.keys() 按照顺序存储特性并且 O(1) 时间复杂度完成存取，可以很好完成题目，并且 get 和 put 时间复杂度都为 O(1)

当前js 代码不能通过所有案例，有大佬能解释一下吗?

/**
 * @param {number} capacity
 */
var LRUCache = function(capacity) {
    this.capacity = capacity;
    this.map = new Map();
};

/** 
 * @param {number} key
 * @return {number}
 */
LRUCache.prototype.get = function(key) {
    let ans = -1;
    if(this.map.has(key)) {
        ans = this.map.get(key);
        this.map.delete(key);
        this.map.set(key, ans);
    }
    return ans;
};

/** 
 * @param {number} key 
 * @param {number} value
 * @return {void}
 */
LRUCache.prototype.put = function(key, value) {
    if(this.map.has(key)) {
        this.map.delete(key);
    }

    if(this.map.size >= this.capacity) {
        let oldkey = this.map.keys().next().value;
        this.map.delete(oldkey);
    }
    this.map.set(key, value);
};

/** 
 * Your LRUCache object will be instantiated and called as such:
 * var obj = new LRUCache(capacity)
 * var param_1 = obj.get(key)
 * obj.put(key,value)
 */