# Leetcode 191 Number of 1 Bits

Source: Internet
Author: User

Leetcode 191 Number of 1 Bits

Solution One (more traditional method): Use to move N to the right, and with 1 think & get 1 of the number; (There is also the use of division/2, obvious division operation efficiency is lower than the displacement)

Time complexity: 0 (LOGN)

`1 intHammingweight (uint32_t N) {2     intCount=0;3     4      while(n!=0) {5         if(N &1) {6++count;7         }8n = n>>1;9     }Ten      One     returncount; A}`

Solution Two:

Methods of using n& (n-1)

Supposing n =0x110101

N n-1 n& (n-1)

step1:110101 110100 110100

step2:110100 110011 110000

step3:110000 101111 100000

step4:100000 011111 000000

It is found that several 1 n& (n-1) get 0, which is significantly more efficient than the previous ones.

Time complexity: O (M), M is the number of 1 in N.

The code is as follows:

`1 intHammingweight (uint32_t N) {2     intCount=0;3     4      while(n!=0) {5count++;6n = n& (n1);7     }8     9     returncount;Ten}`

Leetcode 191 Number of 1 Bits

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