lc
lc2121
hash抽象分组后前缀和
注意 这个地方的下标个数统计要-1...
++(ll)b[i] * (m - i-1);++
class Solution {
typedef long long ll;
public:
vector<long long> getDistances(vector<int>& arr)
{
int n=arr.size();
vector<ll> ret(n);
unordered_map<int,vector<int>> hash;
for(int i=0;i<n;i++)
{
hash[arr[i]].push_back(i);
}
for(auto& [a,b]:hash)
{
int m=b.size();
vector<ll> p(m+1);
for(int i=1;i<=m;i++)
p[i]=p[i-1]+b[i-1];
for(int i=0;i<m;i++)
{
ll f = (ll)b[i] * i - p[i];
ll e = (p[m] - p[i+1]) - ++(ll)b[i] * (m - i-1);++
ret[b[i]] = f + e;
}
}
return ret;
}
};
lc320
优雅的回溯
//缩写
dfs(p + 1, cnt + 1);
//结算 +cnt +p后
dfs(p + 1, 0); // p+1、重置cnt为0
数字串回溯tricks:
++if (cnt > 0)++
++path.erase(path.end() - to_string(cnt).size(), path.end());++
class Solution {
public:
vector<string> generateAbbreviations(string word)
{
int n = word.size();
string path;
vector<string> ret;
auto dfs = [&](this auto&& dfs,int p, int cnt)
{
if (p == n) {
if (cnt > 0)
path += to_string(cnt);
ret.push_back(path);
++if (cnt > 0) path.erase(path.end() - to_string(cnt).size(), path.end());//回溯++
return;
}
//缩写
dfs(p + 1, cnt + 1);
// 结算+保留
if (cnt > 0)
path += to_string(cnt);
path += word[p];
++dfs(p + 1, 0); // p+1、重置cnt为0++
// 回溯
path.pop_back();
if (cnt > 0) {
path.erase(path.end() - to_string(cnt).size(), path.end());
}
};
dfs(0, 0);
return ret;
}
};