zoukankan      html  css  js  c++  java
  • leetcode 762. Prime Number of Set Bits in Binary Representation

    Given two integers L and R, find the count of numbers in the range [L, R] (inclusive) having a prime number of set bits in their binary representation.

    (Recall that the number of set bits an integer has is the number of 1s present when written in binary. For example, 21 written in binary is 10101 which has 3 set bits. Also, 1 is not a prime.)

    Example 1:

    Input: L = 6, R = 10
    Output: 4
    Explanation:
    6 -> 110 (2 set bits, 2 is prime)
    7 -> 111 (3 set bits, 3 is prime)
    9 -> 1001 (2 set bits , 2 is prime)
    10->1010 (2 set bits , 2 is prime)
    

    Example 2:

    Input: L = 10, R = 15
    Output: 5
    Explanation:
    10 -> 1010 (2 set bits, 2 is prime)
    11 -> 1011 (3 set bits, 3 is prime)
    12 -> 1100 (2 set bits, 2 is prime)
    13 -> 1101 (3 set bits, 3 is prime)
    14 -> 1110 (3 set bits, 3 is prime)
    15 -> 1111 (4 set bits, 4 is not prime)
    

    Note:

    1. L, R will be integers L <= R in the range [1, 10^6].
    2. R - L will be at most 10000.

    解法1:

    直接暴力

    class Solution(object):
        def countPrimeSetBits(self, L, R):
            """
            :type L: int
            :type R: int
            :rtype: int
            """
            # for echo num:
            #    count bits in num and judge if it is prime        
            prime_nums = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
            
            def count_1bits(n):
                ans = 0
                while n:
                    ans += 1
                    n = n & (n-1)
                return ans
            
            ans = 0
            for n in range(L, R+1):
                bits = count_1bits(n)
                if bits in prime_nums:
                    ans += 1
            return ans

    解法2:使用dp,比较巧妙!因为 数字num中1的个数=num/2中1的个数+num末尾数字是否为1

    虽然会说超时,但还是值得掌握的。

    class Solution(object):
        def countPrimeSetBits(self, L, R):
            """
            :type L: int
            :type R: int
            :rtype: int
            """      
            prime_nums = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
            
            def count_bits(n):
                bits = [0]*(n+1)
                for i in range(1, n+1):
                    bits[i] = bits[i>>1] + (i&1)
                return bits
    
            ans = 0
            bits = count_bits(R)
            for n in range(L, R+1):
                if bits[n] in prime_nums:
                    ans += 1                
            return ans
  • 相关阅读:
    JDBC存取二进制文件示例
    java多线程向数据库中加载数据
    Lucene建索引代码
    postgresql存储二进制大数据文件
    java项目使用Echarts 做柱状堆叠图,包含点击事件
    子页面获取父页面控件
    JSTL和select标签的组合使用
    log4j配置祥解
    IT项目经理应具备的十大软技能
    Spring和Struct整合的三个方法
  • 原文地址:https://www.cnblogs.com/bonelee/p/8546831.html
Copyright © 2011-2022 走看看