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......
首先用最直观的解法，先将两个binary search tree的元素插入到vector,然后排序。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
......
先用最直白的解法AC：
class Solution {
public:
bool kLengthApart(vector<int>& nums, int k) {
int n = nums.size();
for(int i = 0;i < n;i++)
{
if(nums[i] == 1)
{
int cnt = k;
for(int j = i+1;j < n;j++)
{
if(nums[j] == 1)
{
if(cnt > 0)
return false;
break;
}
if(nums[j] ......
常规暴力破解方法，时间复杂度O(n2)：
class Solution {
public:
vector<int> minOperations(string boxes) {
int n = boxes.size();
vector<int> res;
for(int i = 0;i < n;i++)
{
int distance = 0;
for(int j = 0;j < n;j++)
{
if(i == j) continue;
if('1' == boxes[j])
distance += abs(i-j);
}
res.push_back(d......
工作中有个需要使用sort对从mongo中查询到得数据进行排序，学习了下sort的一些用法。
#include <algorithm>
#include <vector>
bool comp(int d1, int d2)
{
return d1 >= d2;
}
int main()
{
std::vector<int> d(50,10);
std::sort(d.begin(), d.end(), comp);
return 0;
}
上面的代码看似没什么问题，但运行崩溃：
[root workspace]#./sort
**......
递归解法
递归解法，开始解法存在超时，优化方法是将vector的长度放在类成员中或通过参数传递，如果在helper中每次求nums.size()会导致运行超时。
class Solution {
private:
//目标和个数
int count;
//vector大小,减少递归函数里查询O(n)复杂度
int size;
public:
int findTargetSumWays(vector<int>& nums, int S) {
size = nums.size();
helper(nums,0,0,S);
return count;
}
private:......