lc826
++// 记得预处理：维护每个难度对应的最大收益
for(int i=1;i<n;i++){
dpi.second = max(dpi.second, dpi-1.second);++
class Solution {
public:
int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker)
{
vector<pair<int,int>> dp;
int m=worker.size(),n=difficulty.size();
for(int i=0;i<n;i++)
{
dp.push_back({difficultyi,profiti});
}
sort(dp.begin(),dp.end());
++// 预处理：维护每个难度对应的最大收益
for(int i=1;i<n;i++){
dpi.second = max(dpi.second, dpi-1.second);++
}
int ret=0;
for(int i=0;i<m;i++)
{
int t=upper_bound(dp.begin(),dp.end(),make_pair(workeri, INT_MAX))-dp.begin()-1;
if(t>=0) ret+=dpt.second;
}
return ret;
}
};
lcr49
++if(!node->left && !node->right)
{
int t=stoi(path);
ret+=t;
}
else++
class Solution {
public:
int sumNumbers(TreeNode* root)
{
string path;
int ret=0;
auto dfs=\&(this auto&& dfs,TreeNode* node)
{
if(!node)
return;
path+=(node->val+'0');
++if(!node->left && !node->right)
{
int t=stoi(path);
ret+=t;
}
else++
{
dfs(node->left);
dfs(node->right);
}
path.pop_back();
};
dfs(root);
return ret;
}
};
lc1457
mask优化回溯
init 0
- ^(1<<i) 实现标记
__builtin_popcount(mask) <= 1 实现统计
class Solution {
public:
int pseudoPalindromicPaths (TreeNode* root)
{
int mask = 0, ret = 0; // mask用二进制位记录数字出现次数的奇偶
auto dfs=\&(this auto&& dfs,TreeNode* node) {
if(!node) return;
mask ^= (1 << node->val); // 异或：出现次数奇偶翻转
if(!node->left && !node->right) {
// 伪回文：mask中1的数量≤1
ret += (__builtin_popcount(mask) <= 1);
} else {
dfs(node->left);
dfs(node->right);
}
mask ^= (1 << node->val); // 回溯
};
dfs(root);
return ret;
}
};
string 和 unordered_map 效率低了tle了..
++// 仅叶子节点检查路径++
++if(!node->left && !node->right)++
++if(check(path)) ret++;++
class Solution {
public:
int pseudoPalindromicPaths (TreeNode* root)
{
string path;
int ret=0;
auto check=\&(string s) {
unordered_map<char,int> hash;
for(auto& c:s) hashc++;
bool f=false;
for(auto& _,b:hash)
{
if(b%2==1)
{
if(!f) f=true;
else return false;
}
}
return true;
};
auto dfs=\&(this auto&& dfs,TreeNode* node) {
if(!node) return;
path.push_back(node->val+'0');
++// 仅叶子节点检查路径++
++if(!node->left && !node->right) {++
++if(check(path)) ret++;++
}
else {
dfs(node->left);
dfs(node->right);
}
path.pop_back();//回溯
};
dfs(root);
return ret;
}
};
lc1267
预处理统计行列
class Solution {
public:
int countServers(vector<vector<int>>& grid)
{
int m = grid.size(), n = grid0.size();
vector<int> row(m, 0), col(n, 0);
// 先统计每行、每列的服务器数量
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (gridij == 1) {
rowi++;
colj++;
}
}
}
int ret = 0;
// 遍历每个服务器，判断其行/列是否有至少2个服务器
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (gridij == 1 && (rowi > 1 || colj > 1)) {
ret++;
}
}
}
return ret;
}
};