位运算肯定先转二进制，我的快速记忆是都是1反1否则反0
比如 2&1就为10与01进行位运算就为0，3&1就为 11与01进行位运算结果为1
位1的个数https://leetcode-cn.com/problems/number-of-1-bits/submissions/
思路
每次进行消1运算，消一次记录一次
代码
js
var hammingWeight = function (n) {
let res = 0
while (n) {
n &= (n - 1)
res++
}
return res
};
c
int hammingWeight(uint32_t n) {
int res = 0;
while(n){
n &= n-1;
res++;
}
return res;
}
数字范围按位与
https://leetcode-cn.com/problems/bitwise-and-of-numbers-range/
思路因区间在[left, right]，要提取出left与right的公共前缀，思路就是对right进行消一 *** 作当right不再比left大时直接返回right即可
代码
js
var rangeBitwiseAnd = function(left, right) {
while(left
- c
int rangeBitwiseAnd(int left, int right){
while(right > left){
right &= right-1;
}
return right;
}
数组中数字出现的次数 II
https://leetcode-cn.com/problems/shu-zu-zhong-shu-zi-chu-xian-de-ci-shu-ii-lcof/
思路
因为数字有32位，循环32次，每次左移一位，然后每次都与数组的数进行位与 *** 作，看i位是否为1如果为1，计数一次js
var singleNumber = function (nums) {
let res = 0;
for (let i = 0; i < 32; i++) {
let cnt = 0;
let bit = 1 << i;
for (let j = 0; j < nums.length; j++) {
if (nums[j] & bit) cnt++;
}
//找到某一位为1的数，并左移，为了将这一位循环结束后移至原位
if (cnt % 3 != 0) res |= bit;
}
return res;
};
c
int singleNumber(int* nums, int numsSize){
int res = 0;
for(int i = 0; i < 32; i++){
int cnt = 0;
int bit = 1 << i;
for(int j = 0; j < numsSize;j++){
//将数右移，并求出最后一位为 1 的个数
if((nums[j] >> i & 1) == 1) cnt++;
}
if(cnt % 3 != 0)res |= bit;
}
return res;
}
318. 最大单词长度乘积
https://leetcode-cn.com/problems/maximum-product-of-word-lengths/
思路
先将所有字符的bitMask预计算出来，避免重复计算然后判断两个字符串是否重复并取出最大乘积
js
var maxProduct = function (words) {
let n = words.length
let masks = []
for (let i = 0; i < n; i++) {
let bitMask = 0
for (let c of words[i]) {
// 这段代码主要是给字符记位如出现a就标记为1，b标记为10...97为a的字符编码
bitMask |= (1 << c.charCodeAt() - 97)
}
masks[i] = bitMask
}
let ans = 0
for (let i = 0; i < words.length; i++) {
let word1 = words[i]
for (let j = i + 1; j < words.length; j++) {
let word2 = words[j]
// 取得对应的预计算的结果进行位与操作判断是否有重复的字符，并得到最大的单词长度乘积
if ((masks[i] & masks[j]) == 0) ans = Math.max(ans, word1.length * word2.length)
}
}
return ans
}
c
int max(num1,num2){
return num1>num2?num1:num2;
}
int maxProduct(char ** words, int wordsSize){
int masks[wordsSize];
for(int i = 0; i < wordsSize; i++){
int bitMask = 0;
for(int k = 0; k < strlen(words[i]); k++){
if(words[i][k])
if(words[i][k])
bitMask |= (1 << words[i][k] - 'a');
}
masks[i] = bitMask;
}
int ans = 0;
for(int i = 0; i < wordsSize; i++){
int word1 = strlen(words[i]);
for(int j = 0; j < wordsSize; j++){
int word2 = strlen(words[j]);
int num = word1 * word2;
if((masks[i] & masks[j]) == 0) {
ans = max(ans,num);
}
}
}
return ans;
}
1318. 或运算的最小翻转次数
https://leetcode-cn.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/
思路
设a、b和c二进制表示的第i位分别为bit_a、bit_b 和 bit_c若bit_c的值为0，那么bit_a和bit_b必须都为0，需要翻转次数为bit_a+bit_b若bit_c的值为1，那么bit_a和bit_b中至少有一个为1，只有当它们都为0时，才需要1次翻转将每一位的翻转次数进行累加，在枚举完所有位之后，就得到了最小翻转次数
var minFlips = function (a, b, c) {
let ans = 0
for (let i = 0; i < 31; ++i) {
let [bit_a, bit_b, bit_c] = [(a >> i) & 1, (b >> i) & 1, (c >> i) & 1]
if (bit_c == 0) {
ans += bit_a + bit_b
} else {
ans += (bit_a + bit_b == 0)
}
}
return ans
};
c
int minFlips(int a, int b, int c){
int result=0;
for(int i = 0; i < 31; i++){
int bit_a = (a >> i) & 1;
int bit_b = (b >> i) & 1;
int bit_c = (c >> i) & 1;
if(bit_c == 0){
result += bit_a + bit_b;
}else{
result += (bit_a + bit_b == 0);
}
}
return result;
}
461. 汉明距离
https://leetcode-cn.com/problems/hamming-distance/
思路
先进行异或 *** 作，然后消一计数
js
var hammingWeight = function (n) {
let res = 0
while (n) {
n &= (n - 1)
res++
}
return res
};
var hammingDistance = function(x, y) {
let num = y^x;
return hammingWeight(num)
};
c
int hammingWeight(n) {
int res = 0;
while (n) {
n &= (n - 1);
res++;
}
return res;
};
int hammingDistance(int x, int y){
int num = y ^ x;
return hammingWeight(num);
}
136. 只出现一次的数字
https://leetcode-cn.com/problems/single-number/
思路
直接异或最后剩的数就是只出现一次数字的
var singleNumber = function (nums) {
let res = 0;
for (let i = 0; i < nums.length; i++) {
res ^= nums[i];
}
return res;
};
int singleNumber(int* nums, int numsSize){
int s = 0,i;
for(i=0;i
1486. 数组异或 *** 作
https://leetcode-cn.com/problems/xor-operation-in-an-array/
思路
start为循环开始条件，用一个变量来计数，每次循环+=2,并记一次数，当计数为n时循环结束
js
var xorOperation = function (n, start) {
let sum = 0,res = 0;
for (let i = start; sum != n; i += 2, sum++) {
res ^= i
}
return res
};
c
int xorOperation(int n, int start){
int sum = 0,res = 0;
for (int i = start; sum != n; i += 2, sum++) {
res ^= i;
}
return res;
}
1720. 解码异或后的数组
https://leetcode-cn.com/problems/decode-xored-array/
思路
创建一个数组，长度为encoded的长度+1，将第1位为first，再将其他几位做异或 *** 作得到最终数组
js
*/
var decode = function(encoded, first) {
let arr = [first]
for(let [index,item] of encoded.entries()){
arr[index+1] = arr[index] ^ item
}
return arr
};
c
int* decode(int* encoded, int encodedSize, int first, int* returnSize){
int* arr= malloc(sizeof(int) * (encodedSize + 1));
arr[0] = first;
for(int i = 0; i < encodedSize; i++){
arr[i+1] = encoded[i] ^ arr[i];
}
*returnSize = encodedSize + 1;
return arr;
}
1310. 子数组异或查询
https://leetcode-cn.com/problems/xor-queries-of-a-subarray/
思路
首先计数前缀异或数组，然后计算每个的查询结果，第i个就为前缀和异或数组[queries[i][0]]异或前缀和数组[queries[i][1]+1]
js
var xorQueries = function(arr, queries) {
const n = arr.length;
const xors = new Array(n + 1).fill(0);
for (let i = 0; i < n; i++) {
xors[i + 1] = xors[i] ^ arr[i];
}
const m = queries.length;
const ans = new Array(m).fill(0);
for (let i = 0; i < m; i++) {
ans[i] = xors[queries[i][0]] ^ xors[queries[i][1] + 1];
}
return ans;
};
c
int* xorQueries(int* arr, int arrSize, int** queries, int queriesSize, int* queriesColSize, int* returnSize) {
int n = arrSize;
int xors[n + 1];
xors[0] = 0;
for (int i = 0; i < n; i++) {
xors[i + 1] = xors[i] ^ arr[i];
}
int m = queriesSize;
int* ans = malloc(sizeof(int) * m);
*returnSize = m;
for (int i = 0; i < m; i++) {
ans[i] = xors[queries[i][0]] ^ xors[queries[i][1] + 1];
}
return ans;
}
面试题 05.01. 插入
https://leetcode-cn.com/problems/insert-into-bits-lcci/
思路先将i - j位归0，然后将M左移i位进行位或 *** 作
js
var insertBits = function(N, M, i, j) {
let k
for(k = i;k <= j; ++k){
N &= ~(1 << k)
}
return N | (M << i)
};
c
int insertBits(int N, int M, int i, int j){
int k;
for(k = i;k <= j; ++k){
N &= ~((long long)1 << k);
}
return N | (M << i);
}
翻转数位
https://leetcode-cn.com/problems/reverse-bits-lcci/
思路
维护一个窗口，窗口内的0的个数<=1时，窗口大小可以计入结果，取长度的最大值
当窗口内0的个数>1的时候收缩窗口，直到满足<=1
窗口边界的取值范围为[0,31]
js<<
var reverseBits = function (num) {
let s = 0, c = 0, res = -1;
for (let i = 0; i < 32; i++) {
if ((num & (1 << i)) === 0) c += 1;
while (c > 1) {
if ((num & (1 << s)) === 0) c -= 1
s += 1
}
res = Math.max(res, i - s + 1)
}
return res
};
交替位二进制数
https://leetcode-cn.com/problems/binary-number-with-alternating-bits/
思路
判断相邻的两位是否相同，不同就是true，所以从低到高不断位与3判断相邻两位的模式，如果循环结束没有这种模式就返回true
js
var hasAlternatingBits = function(n) {
while(n){
if((n&3) == 3 || (n&3) == 0){
return false
}
n >>= 1;
}
return true
};
c
bool hasAlternatingBits(int n){
while(n){
if((n&3) == 3 || (n & 3) == 0){
return false;
}
n >>= 1;
}
return true;
}
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