convertir este codigo de C a java
Publicado por Vashigo (2 intervenciones) el 10/10/2017 04:04:45
Hola, les agradecería mucho si alguien podría ayudarme
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#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<string> FAIL(1, "fail");
inline char chxor(char ch1, char ch2) {
return (ch1 == '#') ^ (ch2 == '#') ? '#' : '.';
}string invert_line(string s) {
string ret;
for (int i = 0; i+2 < s.size(); i++) {
ret += s[i];
s[i+1] = chxor(s[i], s[i+1]);
s[i+2] = chxor(s[i], s[i+2]);
}if (s[s.size()-2] != '.' || s[s.size()-1] != '.') return "";
return ret;
}vector<string> invert(vector<string> g) {
int sx, sy;for (sy = 0; sy < g.size(); sy++) {
if (invert_line(g[sy]).size() == 0) break;
}if (sy < g.size()) {
for (sx = 0; sx < g[0].size(); sx++) {
string s;
for (int y = 0; y < g.size(); y++) s += g[y][sx];
if (invert_line(s).size() == 0) break;
}if (sx == g[0].size()) return FAIL;
g[sy][sx] = chxor('#', g[sy][sx]);
}vector<string> ret(g.size()-2);
for (int y = 0; y+2 < g.size(); y++) {
ret[y] = invert_line(g[y]);
if (ret[y].size() == 0) return FAIL;
for (int y2 = y+1; y2 <= y+2; y2++)
for (int x = 0; x < ret[y].size(); x++) {
g[y2][x ] = chxor(g[y2][x ], ret[y][x]);
g[y2][x+1] = chxor(g[y2][x+1], ret[y][x]);
g[y2][x+2] = chxor(g[y2][x+2], ret[y][x]);
} }if (g[g.size()-1].find('#') != -1 || g[g.size()-2].find('#') != -1) {
return FAIL;
}int minx = 1000000, maxx = -1, miny = 1000000, maxy = -1;
for (int y = 0; y < ret.size(); y++)
for (int x = 0; x < ret[0].size(); x++) if (ret[y][x] == '#') {
minx = min(minx, x);
maxx = max(maxx, x);
miny = min(miny, y);
maxy = max(maxy, y);
}vector<string> ret2(maxy-miny+1);
for (int y = miny; y <= maxy; y++) {
ret2[y-miny] = ret[y].substr(minx, maxx-minx+1);
}return ret2;
}int main() {
int X, Y;while (cin >> X >> Y) {
vector<string> g(Y);
for (int y = 0; y < Y; y++) cin >> g[y];
while (g.size() >= 3 && g[0].size() >= 3) {
vector<string> g2 = invert(g);
if (g2 == FAIL) break;
g = g2; }for (int y = 0; y < g.size(); y++) cout << g[y] << endl;
}}Valora esta pregunta
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