题目来源：https://leetcode.com/problems/n-queens-ii/
问题描述
52. N-Queens II
Hard
The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
Given an integer n, return the number of distinct solutions to the n-queens puzzle.
Example:
Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],
["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
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题意
N皇后问题。在N×N的棋盘上放N个皇后，使得她们不会互相攻击。与LeetCode 51. N-Queens（dfs）不同的是，这题只要求满足条件的摆放的种数。
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思路
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和LeetCode 51. N-Queens（dfs）完全一样，就是dfs深搜。
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代码
class Solution {
private int ans = 0;
private int n;
private int[] mask;
private boolean[] col_mark;
private boolean[] sub_mark;
private boolean[] sum_mark;
private void init(int n)
{
this.n = n;
this.mask = new int[n];
this.col_mark = new boolean[n];
this.sub_mark = new boolean[2*n-1];
this.sum_mark = new boolean[2*n-1];
}
public void dfs(int x)
{
if (x == n)
{
this.ans++;
return;
}
int y = 0, bit = 1;
for (y=0; y<this.n; y++)
{
if (!this.col_mark[y] && !this.sub_mark[x-y+n-1] && !this.sum_mark[x+y])
{
this.mask[x] = bit;
this.col_mark[y] = true;
this.sub_mark[x-y+n-1] = true;
this.sum_mark[x+y] = true;
dfs(x+1);
this.col_mark[y] = false;
this.sub_mark[x-y+n-1] = false;
this.sum_mark[x+y] = false;
}
bit <<= 1;
}
}
public int totalNQueens(int n) {
if (n == 0)
{
return 1;
}
init(n);
dfs(0);
return ans;
}
}