class Solution {
public:
int singleNumber(vector<int>& nums) {
int size = nums.size();
int ans = 0;
for(int i = 0;i < 32;i++)
{
int sum = 0;
for(auto num:nums)
{
sum += ((num >> i)&1);
}
if(sum % 3)
ans |= (1 << i);
}
return ans;
}
};
运行效率:
Runtime: 4 ms, faster t......
class Solution {
public:
vector<int> singleNumber(vector<int>& nums) {
int sum = 0,flag = 1;
vector<int> ans(2,0);
for(int i = 0;i < nums.size();i++)
sum ^= nums[i];
while(!(sum & flag))
flag <<= 1;
for(int i = 0;i < nums.size();i++)
{
if(flag & nums[i])
an......
安装后修改密码
安装mysql后修改密码
MySQL连接
使用mysql -u root -p 连接，-u指定用户root,-p 密码选项,如果设置了密码需要加-p选项
连接方式一
[root ~]#mysql -u root -proot
mysql: [Warning] Using a password on the command line interface can be insecure.
Welcome to the MySQL monitor. Commands end with ; or \g.
Your MySQL connection id is 10
S......
AC代码：
class Solution {
public:
string multiply(string num1, string num2) {
string res(num1.size()+num2.size(),'0');
for(int i = num1.size() - 1;i >= 0;i--)
{
int carry = 0;
for(int j = num2.size() - 1;j >= 0;j--)
{
int tmp = res[i+j+1]-'0' + (num1[i] - '0')*(num2[j]-'0') + carry;
res[i+j+1]......
首先我们使用一种便于理解的方法，将链表一分为二，然后归并排序，再合并两个有序链表。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* ......
与leetcode15题目类似，大方向是使用双指针 三指针求解问题，细节稍有不同，AC代码如下：
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int n = nums.size();
sort(nums.begin(),nums.end());
int min_val = INT_MAX;
int res = 0;
for(int i = 0;i < n - 2;i++){
int second = i + 1;
int third = n -1;......
方法一：排序
class Solution {
public:
bool canMakeArithmeticProgression(vector<int>& arr) {
sort(arr.begin(),arr.end());
int diff = arr[1] - arr[0];
for(int i = 1;i < arr.size();i++)
{
if(arr[i] - arr[i-1] != diff)
return false;
}
return true;
}
};
运行效率：
Runtime: 4 ms, faster than 65......
class Solution {
public:
string reverseWords(string s) {
int size = s.size();
int i = 0,j = 0;
while(i < size)
{
//skip spaces
while(i < size && s[i] == ' ')
i++;
if(i < size && j > 0)
s[j++] = ' ';
int start = j;
while(i < size && s[i] != ' ')
s[j++] = s[i++];......
比较直观好理解的解法：
class Solution {
public:
int minOperations(int n) {
int res = 0;
for(int i = 1;i <= n;i++)
{
int num = 2*(i-1) + 1;
if(num > n)
res += num -n;
}
return res;
}
};
运行效率:
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Minimum Operations to Make Array Equal......
最直观最容易想的方法，暴力破解方法：
class Solution {
public:
int bulbSwitch(int n) {
vector<int> state(n,0);
for(int i = 1;i <= n;i++)
{
for(int j = i-1;j < n;j += i)
{
state[j] ^= 1;
}
}
return count(state.begin(),state.end(),1);
}
};
执行结果：
Time Limit Exceeded
Details
Last executed input
10000......