N皇后

class Solution {
    private int ans;
    public int totalNQueens(int n) {
        boolean[] col = new boolean[n];
        boolean[] diag1 = new boolean[n*2 - 1];
        boolean[] diag2 = new boolean[n*2 - 1];
        dfs(0, col, diag1, diag2);
        return ans;
    }

    private void dfs(int r, boolean[] col, boolean[] diag1, boolean[] diag2){
        int n = col.length;
        if(r == n){
            ans++;
            return;
        }
        for(int c = 0; c < n; c++){
            int rc = r-c+n-1;//计算当前行r和列c对应的右斜线diag2的索引。
            if(!col[c] && !diag1[r+c]&&!diag2[rc]){
                col[c] = diag1[r+c] = diag2[rc] =true;
                dfs(r+1,col,diag1,diag2);
                col[c] = diag1[r+c] =diag2[rc]=false;
            }
        }
    }
}