题解 | #设计LRU缓存结构#

用有序容器保存hashmap中key的顺序

# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# lru design
# @param operators int整型二维数组 the ops
# @param k int整型 the k
# @return int整型一维数组
#
class Solution:
    def LRU(self , operators: List[List[int]], k: int) -> List[int]:
        # write code here
        templ = []
        tempd = dict()
        answer = []
        for i in operators:
            if i[0] == 1:
                if i[1] in tempd:
                    templ.remove(i[1])
                    templ.insert(0, i[1])
                    tempd[i[1]] = i[2]
                else:
                    if len(templ) < k:
                        tempd[i[1]] = i[2]
                        templ.insert(0, i[1])
                    else:
                        x = templ.pop()
                        templ.insert(0, i[1])
                        tempd.pop(x)
                        tempd[i[1]] = i[2]
            elif i[0] == 2:
                if i[1] in tempd:
                    templ.remove(i[1])
                    templ.insert(0, i[1])
                    answer.append(tempd[i[1]])
                else:
                    answer.append(-1)
        return answer