标签:iso 处理 ons binary orm input nes leetcode desc
Given a positive integer n, find the number of non-negative integers less than or equal to n, whose binary representations do NOT contain consecutive ones.
Input: 5
Output: 5
Explanation:
Here are the non-negative integers <= 5 with their corresponding binary representations:
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule.
1 <= n <= 109
和 902 一样,且边界条件处理要简单的多
class Solution(object):
def findIntegers(self, num):
"""
:type num: int
:rtype: int
"""
ll = [int(i) for i in ‘{0:b}‘.format(num)]
helper = [[0, 0] for i in range(len(ll))]
helper[0][1] = 1
helper[0][0] = 1
if len(helper) > 1:
helper[1][1] = 1
helper[1][0] = 2
for level in range(2, len(ll)):
helper[level][0] = sum(helper[level-1])
helper[level][1] = helper[level-1][0]
ssum = 0
for rlevel, v in enumerate(ll):
level = len(ll) - 1 - rlevel
if v == 0:
if level == 0:
ssum += helper[0][0]
continue
if level > 0:
if ll[rlevel:rlevel+2] == [1, 1]:
return ssum + sum(helper[level])
if level == 0:
if v == 0:
ssum += helper[0][1]
else:
ssum += sum(helper[0])
else:
ssum += helper[level][0]
return ssum
You are here!
Your runtime beats 55.00 % of python submissions.
leetcode: 600. Non-negative Integers without Consecutive Ones
标签:iso 处理 ons binary orm input nes leetcode desc
原文地址:https://www.cnblogs.com/tmortred/p/13118097.html