LeetCode "Strobogrammatic Number III"

It can be solved based on the code from "Strobogrammatic Number II". The idea is pretty straight forward - binary search the boundaries.

class Solution
{
int go_cnt(int n, bool bInner)
{
int ret = 0;
switch (n)
{
case 0:
ret = 1; // ""
break;
case 1:
ret = 3; //{ "0", "1", "8" };
break;
case 2:
if(bInner) ret += 1; //"00"
ret += 4;    //{ "11", "69", "88", "96" };
break;
default:
int prev = go_cnt(n - 2, true);
int fac = 0;
if(bInner) fac += 1;
fac += 4;

ret = fac * prev;
break;
}
return ret;
}

vector<string> go(int n, bool bInner)
{
vector<string> ret;
if (n == 0) return ret;
if (n == 1)
{
ret.push_back("0");
ret.push_back("1");
ret.push_back("8");
}
else if (n == 2)
{
if (bInner)
{
ret.push_back("00");
}
ret.push_back("11");
ret.push_back("69");
ret.push_back("88");
ret.push_back("96");
}
else
{
vector<string> prev = go(n - 2, true);
if(bInner)
{
for (auto &s : prev)
ret.push_back('0' + s + '0');
}
for (auto &s : prev)
ret.push_back("1" + s + "1");
for (auto &s : prev)
ret.push_back("6" + s + "9");
for (auto &s : prev)
ret.push_back("8" + s + "8");
for (auto &s : prev)
ret.push_back("9" + s + "6");
}
return ret;
}
public:
int strobogrammaticInRange(string low, string high)
{
int ret = 0;
int lenl = low.length(), lenh = high.length();
if(lenl > lenh) return 0;

//    complete
for(int j = lenl + 1; j < lenh; j ++)
ret += go_cnt(j, false);

int cnt0 = 0, cnt1 = 0;
auto r0 = go(lenl, false);

auto it0 = std::lower_bound(r0.begin(), r0.end(), low);
cnt0 = r0.end() - it0;

if(lenh != lenl)
{
auto r1 = go(lenh, false);
auto it1 = std::upper_bound(r1.begin(), r1.end(), high);
cnt1 = it1 - r1.begin();

ret += cnt0 + cnt1;
}
else
{
auto it1 = std::upper_bound(r0.begin(), r0.end(), high);
cnt1 = it1 - r0.begin();

ret = cnt0 + cnt1 - r0.size();
}

return ret;
}
};

View Code
Of course, we can optimize the vector generation part: we only need one part of it :)