LeetCode - N-Queens II

N-Queens II

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 };