启动redis
[root ~]#redis-server
8988:C 25 Jun 11:24:31.317 # Warning: no config file specified, using the default config. In order to specify a config file use redis-server /path/to/redis.conf
8988:M 25 Jun 11:24:31.319 * Increased maximum number of open files to 10032 (it was originally set to 1024).......
class Solution {
public:
int firstMissingPositive(vector<int>& nums) {
int n = nums.size();
for(int i = 0;i < n;i++)
{
if(nums[i] <= 0)
nums[i] = n + 1;
}
for (int i = 0; i < n; ++i) {
int num = abs(nums[i]);
if (num <= n) {
nums[num - 1] = -abs(nums[num - 1]);
}
}
for(int i......
class Solution {
public:
int compareVersion(string version1, string version2) {
istringstream v1(version1);
istringstream v2(version2);
while(!v1.eof() || !v2.eof())
{
string s1 = "0",s2 = "0";
if(!v1.eof())
getline(v1,s1,'.');
if(!v2.eof())
getline(v2,s2,'.');
int n1 = atoi(s1.c_str());
int n2 =......
class Solution {
public:
int divide(int dividend, int divisor) {
if(dividend == INT_MIN && divisor == -1) return INT_MAX;
long long A = llabs(dividend),B = llabs(divisor),ans = 0;
while(A >= B)
{
long long base = B,m = 1;
while(base << 1 <= A)
{
base <<= 1;
m <<= 1;
......
冒泡排序
class Solution {
public:
vector<int> sortArray(vector<int>& nums) {
for(int i = nums.size()-1;i > 0;i--)
{
for(int j = 0;j < i;j++)
if(nums[j] > nums[j+1])
swap(nums[j],nums[j+1]);
}
return nums;
}
};
执行结果：
Time Limit Exceeded
Details
Last executed input
[5864,-1......
#include <semaphore.h>
class Foo {
protected:
sem_t firstJobDone;
sem_t secondJobDone;
public:
Foo() {
sem_init(&firstJobDone, 0, 0);
sem_init(&secondJobDone, 0, 0);
}
void first(function<void()> printFirst) {
// printFirst() outputs "first". Do not change or remove th......
输入一个整数 n ，求1～n这n个整数的十进制表示中1出现的次数。
例如，输入12，1～12这些整数中包含1 的数字有1、10、11和12，1一共出现了5次。
class Solution {
public:
int countDigitOne(int n) {
int res = 0;
for(int i = 1;i <= n;i++)
{
res += helper(i);
}
return res;
}
private:
int helper(int n)
{
int cnt = 0;
while(n)
{
if(n%10 == 1)
cnt++;
n /= 10;
}
r......
方法一:记录每个字符出现次数
class Solution {
public:
char firstUniqChar(string s) {
unordered_map<char,int> hash;
for(auto c:s)
hash[c] += 1;
for(int i = 0;i < s.size();i++)
{
if(hash[s[i]] == 1)
return s[i];
}
return ' ';
}
};
方法二：
class Solution {
public:
char firstUniqChar(string s) {
unorde......
递归解法
class Solution {
public:
int integerBreak(int n) {
if(n <= 2) return 1;
int ans = INT_MIN;
for(int i = 2;i < n;i++)
{
int product = i * max(n-i,integerBreak(n-i));
if(product > ans)
ans = product;
}
return ans;
}
};
运行结果:
Time Limit Exceeded
Details
Last executed input
43
......
自己实现的方法一，思路是先获取空格的个数，然后根据空格个数申请新的字符串长度。
class Solution {
public:
string replaceSpace(string s) {
int size = s.size();
int cnt = 0,loc = 0;
for(int i = 0;i < size;i++)
if(s[i] == ' ')
cnt++;
string res(size + 2*cnt,'0');
for(int i = 0;i < size;i++)
{
if(s[i] != ' ')
{
......