Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.
把051的回溯代码改一下,把添加结果的代码改为结果数目加1即可。
class Solution {
public:
int totalNQueens(int n) {
string s(n, '.');
vector<string> result(n, s);
results_num = 0;
solve(result, 0, n);
return results_num;
}
void solve(vector<string> &result, int index, int n) {
if (index == n) {
results_num++;
}
for (int j = 0; j < n; j++) {
if (check(result, index, j, n)) {
result[index][j] = 'Q';
solve(result, index+1, n);
result[index][j] = '.';
}
}
}
bool check(vector<string> &result, int i, int j, int n) {
int x = i-1;
int y = j;
int y1 = j-1;
int y2 = j+1;
while (x >= 0) {
if (result[x][y] == 'Q') {
return false;
}
if (y1 >= 0 && result[x][y1] == 'Q') {
return false;
}
if (y2 < n && result[x][y2] == 'Q') {
return false;
}
x--;
y1--;
y2++;
}
return true;
}
/* challenge
bool check(vector<string>& result, int i, int j, int n) {
for (int row = 1; row <= i; ++row) {
if (result[i - row][j] == 'Q' || (j - row >= 0 && result[i - row][j - row] == 'Q') ||
(j + row < n && result[i - row][j + row] == 'Q')) {
return false;
}
}
return true;
}
*/
private:
int results_num;
};