leetcode1387

class Solution:
    def getKth(self, lo: int, hi: int, k: int) -> int:
        memo = dict()
        memo[1] = 0
        for i in range(1,hi+1):
            cur = i
            count = 0
            while cur != 1:
                count += 1
                if cur % 2 == 0:#even
                    cur = cur // 2
                else:
                    cur = 3 * cur + 1
                if cur in memo:
                    count += memo[cur]
                    break
            memo[i] = count
        #print(memo)
        temp = []
        for key,val in memo.items():
            if key >= lo and key <= hi:
                temp.append((key,val))
        result = sorted(temp,key=lambda x:(x[1],x[0]))
        return result[k-1][0]

算法思路：先使用备忘录机制(缓存)，计算从1到hi的所有数字的steps。

然后按照题目要求，进行双条件排序。