52. N-Queens II #
题目 #
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.."]
]
题目大意 #
给定一个整数 n，返回 n 皇后不同的解决方案的数量。
解题思路 #
- 这一题是第 51 题的加强版，在第 51 题的基础上累加记录解的个数即可。
- 这一题也可以暴力打表法，时间复杂度为 O(1)。
代码 #
package leetcode
// 解法一，暴力打表法
func totalNQueens(n int) int {
res := []int{0, 1, 0, 0, 2, 10, 4, 40, 92, 352, 724}
return res[n]
}
// 解法二，DFS 回溯法
func totalNQueens1(n int) int {
col, dia1, dia2, row, res := make([]bool, n), make([]bool, 2*n-1), make([]bool, 2*n-1), []int{}, 0
putQueen52(n, 0, &col, &dia1, &dia2, &row, &res)
return res
}
// 尝试在一个n皇后问题中, 摆放第index行的皇后位置
func putQueen52(n, index int, col, dia1, dia2 *[]bool, row *[]int, res *int) {
if index == n {
*res++
return
}
for i := 0; i < n; i++ {
// 尝试将第index行的皇后摆放在第i列
if !(*col)[i] && !(*dia1)[index+i] && !(*dia2)[index-i+n-1] {
*row = append(*row, i)
(*col)[i] = true
(*dia1)[index+i] = true
(*dia2)[index-i+n-1] = true
putQueen52(n, index+1, col, dia1, dia2, row, res)
(*col)[i] = false
(*dia1)[index+i] = false
(*dia2)[index-i+n-1] = false
*row = (*row)[:len(*row)-1]
}
}
return
}
// 解法三 二进制位操作法
// class Solution {
// public:
// int totalNQueens(int n) {
// int ans=0;
// int row=0,leftDiagonal=0,rightDiagonal=0;
// int bit=(1<<n)-1;//to clear high bits of the 32-bit int
// Queens(bit,row,leftDiagonal,rightDiagonal,ans);
// return ans;
// }
// void Queens(int bit,int row,int leftDiagonal,int rightDiagonal,int &ans){
// int cur=(~(row|leftDiagonal|rightDiagonal))&bit;//possible place for this queen
// if (!cur) return;//no pos for this queen
// while(cur){
// int curPos=(cur&(~cur + 1))&bit;//choose possible place in the right
// //update row,ld and rd
// row+=curPos;
// if (row==bit) ans++;//last row
// else Queens(bit,row,((leftDiagonal|curPos)<<1)&bit,((rightDiagonal|curPos)>>1)&bit,ans);
// cur-=curPos;//for next possible place
// row-=curPos;//reset row
// }
// }
// };