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command + shift + [ 回退
command + shift + ] 前进
command + O 按两次shift键 查找符号 文件 类等
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......
最直观的解法：
class Solution {
public:
int countMatches(vector<vector<string>>& items, string ruleKey, string ruleValue) {
int index = (ruleKey == "type")? 0 : (ruleKey == "color") ? 1: 2;
int res = 0;
for(int i = 0;i < items.size();i++)
{
if(items[i][index] == ruleValue)
res += 1;
}......
先使用最直观的解法：
class CustomStack {
public:
CustomStack(int maxSize) {
m_maxsize = maxSize;
}
void push(int x) {
if(m_elem.size() >= m_maxsize)
return;
m_elem.push(x);
}
int pop() {
if(m_elem.empty())
return -1;
int x = m_elem.top();
m_elem.pop();
return x;
}
void increment(int k, int val) {
sta......