publicinthammingWeight2(intn){intcount =0;for(inti=0; i<32; i++) { count += n&1; n = n >>1; }returncount; } 04 第三种解法 上面的两种解法都是拿n和1做与运算,其实还可以换种思路,使用n和n-1做位运算。此时,分两种情况:当n的二进制最后一位是1的时候,n-1的二进制位除了最后一位...
publicinthammingWeight3(intn){intcount=0;while(n!=0){count++;n=n&(n-1);}returncount;} 05 第四种解法 将n变成二进制字符串,然后依次查找字符串中的1,并记数。其中,整数1的Unicode十进制值是49,所以使用charAt方法时,是和49判断。 publicinthammingWeight4(intn){intcount=0;String binary=Integer....
Leet Code之Number of 1 bits 常规方法,一位一位右移,算出1的个数。但是最大的数有32个1,需要右移32次,效率较低。 常规方法: 1inthammingWeight(uint32_t n) {2intk,num;3num =0;4while(n!=0)5{6k = n%2;7if(k)num++;8n = n>>1;9}10returnnum;11} 效率较高方法:通过将n与n-1进行...
publicinthammingWeight(intn){n=(n&0x55555555)+((n>>>1)&0x55555555);// 32 组向 16 组合并,合并前每组 1 个数n=(n&0x33333333)+((n>>>2)&0x33333333);// 16 组向 8 组合并,合并前每组 2 个数n=(n&0x0f0f0f0f)+((n>>>4)&0x0f0f0f0f);// 8 组向 4 组合并,合并前每组 4...
A new approach, which uses histogram technique along with least square error minimization technique for determination of effective number of bits of an ADC is presented. Histogram technique is used for computation of ADC code bin and transfer characteristic in actual conditions. Least square error ...
bits of an n bit word; and a 1/0 Bitzhlereinheit (16) is associated with n inputs, each one of the outputs of the I / O latch and to generate m output signals to asynchronously within a single cycle of these m outputs, the number of 1 / represent 0 bits in an n-bit word. ...
[LeetCode]Number of 1 Bits Question Write a function that takes an unsigned integer and returns the number of ’1’ bits it has (also known as the Hamming weight). For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function...
Write a function that takes an unsigned integer and returns the number of ’1’ bits it has (also known as the Hamming weight). For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function should return 3. ...
191. Number of 1 Bits 题目: https://leetcode.com/problems/number-of-1-bits/ 难度: Easy 转成二进制,数1的个数
bitcount += 1; value = value & (value - 1); } return bitcount; }It's a silly trick, but clever.Btw, being geeks, a number of people over here came up with the fastest known version (in software). It involves adding the bits in parallel - you can ...