2014.2.13 20:01
Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.
Solution:
This problem is a simplification from the N-Queens. This time we only have record the number of solutions.
Time complexity is O(n!). Space complexity is O(n!) as well, which comes from parameters in recursive function calls.
Accepted code:
1 // 3CE, 1AC, why so hasty? 2 class Solution { 3 public: 4 int totalNQueens(int n) { 5 a = nullptr; 6 if (n <= 0) { 7 return 0; 8 } 9 10 res_count = 0; 11 a = new int[n]; 12 solveNQueensRecursive(0, a, n); 13 delete[] a; 14 15 return res_count; 16 } 17 private: 18 int *a; 19 int res_count; 20 21 void solveNQueensRecursive(int idx, int a[], const int &n) { 22 if (idx == n) { 23 // one solution is found 24 ++res_count; 25 return; 26 } 27 28 int i, j; 29 // check if the current layout is valid. 30 for (i = 0; i < n; ++i) { 31 a[idx] = i; 32 for (j = 0; j < idx; ++j) { 33 if (a[j] == a[idx] || myabs(idx - j) == myabs(a[idx] - a[j])) { 34 break; 35 } 36 } 37 if (j == idx) { 38 // valid layout. 39 solveNQueensRecursive(idx + 1, a, n); 40 } 41 } 42 } 43 44 int myabs(const int x) { 45 return (x >= 0 ? x : -x); 46 } 47 };