600. Non-negative Integers without Consecutive Ones
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.
Example 1:
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.
Note: 1 <= n <= 109
Approach #1: DP. [C++]
class Solution {
public:
int findIntegers(int num) {
vector<int> f(35, 0);
f[0] = 1;
f[1] = 2;
for (int i = 2; i < 32; ++i)
f[i] = f[i-1] + f[i-2];
int ans = 0, k = 30, pre_bit = 0;
while (k >= 0) {
if (num & (1 << k)) {
ans += f[k];
if (pre_bit == 1) return ans;
pre_bit = 1;
} else pre_bit = 0;
k--;
}
return ans+1;
}
};
Analysis:
The solution if based on 2 fact:
First: the number of length k string without consecutive 1 is Fibonacci sequence f(k);
For example, is k = 5, the range is 00000 - 11111. We can consider it as two ranges, which are 00000 - 01111 ans 10000 10111. any number >= 11000 is not allowed due to consecutive 1. The first case is actually f(4), and the second case is f(3), so f(5) = f(4) + f(3).
Second: Scan the number from most significant digit, i.e. left to right, in binary format. If we find a '1' with k digits to the right, count increases by f(k) beause we can put a '0' at this digit and any valid length k string behind; After that, we continue the loop to consider the remaining case, i.e. we put a '1' at this digit. If consecutive 1s are found, we exit the loop and return the answer. By the end of the loop, we return ans + 1 to include the number n itself.
Reference:
https://leetcode.com/problems/non-negative-integers-without-consecutive-ones/discuss/103754/C%2B%2B-Non-DP-O(32)-Fibonacci-solution
600. Non-negative Integers without Consecutive Ones的更多相关文章
- Non-negative Integers without Consecutive Ones
n位二进制,求不包含连续1的二进制(n位)数字个数. http://www.geeksforgeeks.org/count-number-binary-strings-without-consecut ...
- 第十六周 Leetcode 600. Non-negative Integers without Consecutive Ones(HARD) 计数dp
Leetcode600 很简单的一道计数题 给定整数n 求不大于n的正整数中 二进制表示没有连续的1的数字个数 在dp过程中只要保证不出现连续1以及大于n的情况即可. 所以设计按位dp[i][j]表示 ...
- [LeetCode] Non-negative Integers without Consecutive Ones 非负整数不包括连续的1
Given a positive integer n, find the number of non-negative integers less than or equal to n, whose ...
- [Swift]LeetCode600. 不含连续1的非负整数 | Non-negative Integers without Consecutive Ones
Given a positive integer n, find the number of non-negativeintegers less than or equal to n, whose b ...
- [Algorithm] Count Negative Integers in Row/Column-Wise Sorted Matrix
// Code goes here function countNegative (M, n, m) { count = ; i = ; j = m - ; && i < n) ...
- 【LeetCode】动态规划(下篇共39题)
[600] Non-negative Integers without Consecutive Ones [629] K Inverse Pairs Array [638] Shopping Offe ...
- Interleaving Positive and Negative Numbers
Given an array with positive and negative integers. Re-range it to interleaving with positive and ne ...
- [LintCode] Interleaving Positive and Negative Numbers
Given an array with positive and negative integers. Re-range it to interleaving with positive and ne ...
- Lintcode: Interleaving Positive and Negative Numbers 解题报告
Interleaving Positive and Negative Numbers 原题链接 : http://lintcode.com/zh-cn/problem/interleaving-pos ...
随机推荐
- Spring整合Struts2框架的第一种方式(Action由Struts2框架来创建)。在我的上一篇博文中介绍的通过web工厂的方式获取servcie的方法因为太麻烦,所以开发的时候不会使用。
1. spring整合struts的基本操作见我的上一篇博文:https://www.cnblogs.com/wyhluckdog/p/10140588.html,这里面将spring与struts2 ...
- sql标量值函数,将汉字转化为拼音,无音标
USE [db_Test]GO SET ANSI_NULLS ONGO SET QUOTED_IDENTIFIER ONGO create function [dbo].[fn_GetPinyin]( ...
- CF 1023D Array Restoration - 线段树
题解 非常容易想到的线段树, 还可以用并查集来. 还有一位大神用了$O(n)$ 就过了Orz 要判断是否能染色出输入给出的序列,必须满足两个条件: 1. 序列中必须存在一个$q$ 2. 两个相同的数$ ...
- Python图像处理库:Pillow 初级教程-乾颐堂
Image类 Pillow中最重要的类就是Image,该类存在于同名的模块中.可以通过以下几种方式实例化:从文件中读取图片,处理其他图片得到,或者直接创建一个图片. 使用Image模块中的open函数 ...
- 如何从dvi生成pdf--------亲测有效果.
用里面第二个命令. http://blog.csdn.net/u014682350/article/details/46482477
- yii2 ActiveRecord的生命周期
AR的生命周期 http://www.yii-china.com/doc/guide/db_active_record.html 理解AR的生命周期对于你操作数据库非常重要.生命周期通常都会有些典型的 ...
- Django入门指南-第7章:模板引擎设置(完结)
http://127.0.0.1:8000/ <!--templates/home.html--> <!DOCTYPE html> <html> <head& ...
- newton法分形图
方程:z^6-1=0; %f为求解的方程,df是导数,使用的时候用funchandler定义 %res是目标分辨率,iter是循环次数,(xc,yc)是图像的中心,xoom是放大倍数 %参数视自己需求 ...
- Python 析构方法__del__
class Car: def __init__(self): print('---ok---') def __del__(self): print('----deconstrcut-------') ...
- MFC框架仿真<三>R T T I
RTTI,简单的说,就是判定A类是否为B类的基类.将书本的内容最大程度的简化,如下图的类层次,现在解决的问题就是:判定“梨”是否是“红富士”的基类.