LeetCode

191 - Number of 1 Bits

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.

剑指offer上给的解法，每次&(n-1)可以将最右端的1变为0，且只会将最右端的1变为0，循环操作下去可以求出有多少个1

class Solution {public:    int hammingWeight(uint32_t n) {        int ans = 0;        while (n) {            n &= (n - 1);            ans++;        }        return ans;    }};

190. Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

Follow up:
If this function is called many times, how would you optimize it?

Related problem: Reverse Integer

跟上边那题思路是一样的，如果不想改变n的值的话，可以写在循环里  ((n >> i) & 1)

class Solution {public:    uint32_t reverseBits(uint32_t n) {        int ans = 0;        for (int i = 0; i < 32; ++i, n >>= 1) {            ans <<= 1;            ans += n & 1;        }        return ans;    }};

338. Counting Bits

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example:
For num = 5 you should return [0,1,1,2,1,2].

Follow up:

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
• Space complexity should be O(n).
• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

稍微有点难度？求出从0~num所有的数的二进制有多少个1，返回一个数组

本来分析了一下，是个有规律增加的情况，然后写了一波

000 001 010 011

100 101 110 111      

规律：发现只有第一位改变了，后面都没有变，然后推广到更大的数

AC后看了看评论区，嗯。。。果然在下还是不够机智orzzzzzzz

时间复杂度都是O(n)，空间复杂度也都是O(n)

class Solution {    //自己强行找规律public:    vector<int> countBits(int num) {        vector<int> ans(num+1, 0);        ans[0] = 0;        ans[1] = 1;        ans[2] = 1;        ans[3] = 2;        int cnt = 0, cur = 4;        for (int i = 4; i <= num; i++) {            ans[i] = ans[cnt++] + 1;            if (cnt == cur) {                cnt = 0;                cur *= 2;            }        }        return ans;    }};

i & (i - 1) 可得到小于i且二进制1比i少一位的数，然后在此基础上+1就好

class Solution {      //评论区QAQpublic:    vector<int> countBits(int num) {        vector<int> ret(num+1, 0);        for (int i = 1; i <= num; ++i)            ret[i] = ret[i&(i-1)] + 1;   //举一反三路漫漫兮啊，叹气        return ret;    }};