# N-Queens II

Now, instead outputting board configurations, return the total number of distinct solutions.

## Solution

1. Recursion.
2. Recursion + bit version. (fast) The idea is from http://www.matrix67.com/blog/archives/266 (in chinese).
3. Iteration.

## Code

``````public class Solution {
public int totalNQueens(int n) {
}
public int totalNQueens_1(int n) {
int[] board = new int[n];
Arrays.fill(board,-1);
int[] res = new int[1];
totalNQueensRe(n, 0, board, res);
return res[0];
}
public void totalNQueensRe(int n, int row, int[] board, int[] res) {
if (n == row) {
res[0]++;
return;
}
for (int i = 0; i < n; ++i) {
if (isValid(board, row, i)) {
board[row] = i;
totalNQueensRe(n, row + 1, board, res);
board[row] = -1;
}
}
}
public boolean isValid(int[] board, int row, int col) {
for (int i = 0; i < row; ++i) {
if (board[i] == col || row - i == Math.abs(col - board[i]))
return false;
}
return true;
}
public int totalNQueens_2(int n) {
int[] res = new int[1];
totalNQueensRe2(n, 0, 0, 0, res);
return res[0];
}
public void totalNQueensRe2(int n, int row, int ld, int rd, int[] res) {
if (row == (1<<n) -1 ) {
res[0]++;
return;
}
int avail = ~(row | ld | rd);
for (int i = n - 1; i >= 0; --i) {
int pos = 1<<i;
if ((int)(avail&pos) != 0) {
totalNQueensRe2(n, row | pos, (ld|pos) << 1, (rd|pos) >>1, res);
}
}
}
public int totalNQueens_3(int n) {
int[] a = new int[n];
Arrays.fill(a,-1);
int res = 0;
int row = 0;
while (row >= 0) {
if (row == n) {
res++; row--;
}
int i = a[row] == -1 ? 0 : a[row] + 1;
for ( ; i < n; ++i) {
if (isValid(a, row, i)) {
a[row] = i;
row++;
break;
}
}
if (i == n) {
a[row] = -1;
row--;
}
}
return res;
}
}
``````