题目
简单题
剑指 Offer II 003. 前 n 个数字二进制中 1 的个数
中等题
371. 两整数之和 代码是真简单，想是真的难想 秒啊
477. 汉明距离总和 咦 会超时。。。
难题
67. 二进制求和 尝试不去使用字符转整形和整形转字符 只使用字符加法？
260. 只出现一次的数字 III 想想分类
题解
136. 只出现一次的数字 异或的性质
int singleNumber(int* nums, int numsSize){
int ans = 0;// 0和任何数异或还是0
for(int i = 0;i < numsSize;i++)
ans ^= nums[i];
return ans;
}
bool getbit(uint32_t n,int k){
return n & ((uint32_t)1<<k);
}
uint32_t reverseBits(uint32_t n) {
for(int i = 0; i < 16; ++i)
if(getbit(n,i) != getbit(n, 31-i)){
n ^= (uint32_t)1 << i;
n ^= (uint32_t)1 << 31 - i;
}
return n;
}
268. 丢失的数字异或的性质
int missingNumber(int* nums, int numsSize){
unsigned ans = 0;
for(int i = 0;i < numsSize;++i)//全体的异或
ans ^= nums[i];
for(int i = 0;i < numsSize + 1;i++)//0-n的异或
ans ^= i;
return ans;
}
137. 只出现一次的数字 II
int singleNumber(int* nums, int numsSize){
unsigned ans = 0;
for(unsigned i = 0;i < 32;i++){
unsigned temp = 0;
for(int j = 0;j < numsSize;j++)
temp += (((unsigned) 1<<i)&nums[j])>>i; //计算此位所有元素的和
temp %= 3;//出现三次就是能被3整除 结果只会是0或者1
ans += temp << i;
}
return ans; //强制类型转换
}
338. 比特位计数
int* countBits(int n, int* returnSize){
*returnSize = n + 1;
int *ans = malloc(sizeof(int) * (n + 1)),size= 0;//返回数组
for(int i = 0;i <= n;i++){//遍历
int temp = i,count = 0;
while(temp){//计数
if(temp & 1) count++;
temp >>= 1;
}
ans[size++] = count;
}
return ans;
}
389. 找不同
char findTheDifference(char * s, char * t){
char ans = 0;
for(int i = 0;s[i];i++)
ans ^= s[i];
for(int j = 0;t[j];++j)
ans ^= t[j];
return ans;
}
int countbit(int n){//数个数
int count = 0;
while(n){
if(n&1) count++;
n>>=1;
}
return count;
}
char ** readBinaryWatch(int turnedOn, int* returnSize){
char** ans = malloc(sizeof(char*) * 12 * 60); //申请空间 最多60种
*returnSize = 0; //返回大小
for (int h = 0; h < 12; ++h) {//枚举
for (int m = 0; m < 60; ++m) {
if (countbit(h) +countbit(m) == turnedOn) {
char* tmp = malloc(sizeof(char) * 6); //申请返回空间 6个字符刚好
sprintf(tmp, "%d:%02d", h, m);
ans[(*returnSize)++] = tmp;
}
}
}
return ans;
}
461. 汉明距离
int hammingDistance(int x, int y){
int count = 0;
for(unsigned i = 1;i < 2147483648;i<<=1)//计算距离
if((i&x)^(i&y)) count++;
return count;
}
int subsetXORSum(int* nums, int numsSize){
int num = 0;
for(int i = 0; i < (1 << numsSize); ++i){//所有子集
int ans = 0;
for(int j = 0; j < numsSize; ++j)//求异或
if(i&(1<<j))
ans ^= nums[j];
num += ans;//加和
}
return num;
}
剑指 Offer II 003. 前 n 个数字二进制中 1 的个数
int* countBits(int n, int* returnSize){
*returnSize = n + 1;
int *ans = malloc(sizeof(int) * (n + 1)),size= 0;
for(int i = 0;i <= n;i++){
int temp = i,count = 0;
while(temp){
if(temp & 1) count++;
temp >>= 1;
}
ans[size++] = count;
}
return ans;
}
int* decode(int* encoded, int encodedSize, int first, int* returnSize){
*returnSize = encodedSize + 1;
int *ans = malloc(sizeof(int)*(encodedSize + 1) );
ans[0] = first;
for(int i = 0; i < encodedSize; ++i)
ans[i+1] = encoded[i] ^ ans[i];
return ans;
}
78. 子集
int** subsets(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
*returnSize = pow(2,numsSize);
(*returnColumnSizes) = malloc(sizeof(int)*(*returnSize));
int **ans = malloc(sizeof(int *)*(*returnSize));//申请指针空间
int row = 0;
ans[row] = 0;
(*returnColumnSizes)[row++] = 0;
for(int i = 1;i < pow(2,numsSize);i++){//循环所有生成所有子集
int *hash = malloc(sizeof(int) * 10),hashnum = 0;//申请每行的空间
for(int j = 0;j <10;j++)
if((1<<j)&i) hash[hashnum++] = nums[j];
ans[row] = hash;
(*returnColumnSizes)[row++] = hashnum;
}
return ans;
}
371. 两整数之和
int getSum(int a, int b){
while(b){
unsigned temp = ((unsigned)a&(unsigned)b) << 1;
a = a ^ b;
b = temp;
}
return a;
}
int totalHammingDistance(int* nums, int numsSize){
int ans = 0;
for(int i = 0;i < 32;i++){//按位计算
int count = 0;
for(int j = 0;j < numsSize;j++)
if(nums[j] & ((unsigned)1<<i)) count++;
ans+=(numsSize - count)*count;//此位贡献的海明距离
}
return ans;
}
int* xorQueries(int* arr, int arrSize, int** queries, int queriesSize, int* queriesColSize, int* returnSize){
int *ans=malloc(sizeof(int)*(queriesSize));
int qianzui[arrSize + 1];//前缀异或
*returnSize = 0;
qianzui[0] = 0;
for(int i = 0;i<arrSize;i++)
qianzui[i+1] = qianzui[i] ^arr[i];
for(int i = 0;i <queriesSize;i++)
ans[(*returnSize)++] = qianzui[queries[i][0]] ^ qianzui[queries[i][1]+1];
return ans;
}
二进制求和 尝试不去使用字符转整形和整形转字符 只使用字符加法？
char * addBinary(char * a, char * b){
char *ans = malloc(sizeof(char)*10010);
int ansnum = 0;
int jinwei = 0; //进位的值
int i = strlen(a) - 1,j = strlen(b) - 1;
for(;i >=0 && j >= 0;i--,j--){ //处理字符串
if(a[i] -'0' + b[j] - '0' + jinwei >= 2){ //产生进位
ans[ansnum++] = a[i] - '0' + b[j] + jinwei - 2;//记录结果
jinwei = 1;
}
else{
ans[ansnum++] = a[i] - '0' + b[j] +jinwei;
jinwei = 0;
}
}
while(i >= 0) { //处理a过长
if(a[i] - '0' + jinwei >= 2){
ans[ansnum++] = a[i] +jinwei - 2;
jinwei = 1;
}
else{
ans[ansnum++] = a[i] + jinwei;
jinwei = 0;
}
i--;
}
while(j >= 0) { //处理b过长
if(b[j] - '0' + jinwei >= 2){
ans[ansnum++] = b[j] +jinwei - 2;
jinwei = 1;
}
else{
ans[ansnum++] = b[j] + jinwei;
jinwei = 0;
}
j--;
}
if(jinwei) ans[ansnum++] = jinwei +'0'; //处理最终进位
for(int i = 0;i < ansnum/2;i++){ //反转字符串
ans[i] = ans[i] ^ ans[ansnum - i - 1];
ans[ansnum - i - 1] = ans[i] ^ ans[ansnum - i - 1];
ans[i] = ans[i] ^ ans[ansnum - i - 1];
}
ans[ansnum] = 0;
return ans;
}
260. 只出现一次的数字 III 想想分类
int* singleNumber(int* nums, int numsSize, int* returnSize){
unsigned ans = 0, *ans1= malloc(sizeof(unsigned)*2);
memset(ans1,0,sizeof(ans1));
unsigned i ;
for(i = 0;i <numsSize;++i)
ans ^= nums[i]; //求出a^b
for(i = 1;i < 4294967296;i<<=1)
if(i & ans) break; //第i位不同
for(int j = 0;j < numsSize;++j)
if(nums[j] & i) ans1[0] ^= nums[j];
else ans1[1] ^= nums[j];
*returnSize = 2;
return ans1;
}