福哥答案 2021-02-16：
天然智慧即可。
1. 一般递归。有代码。
须要判断同列和斜线。
2. 位运算递归。有代码。
3. 我的递归。有代码。
只须要判断斜线。
代码用 golang 编写，代码如下：
package main
import (
"fmt"
"time"
)
func main() {
n := 12
fmt.Println(n, "皇后问题")
fmt.Println("------")
now := time.Now()
fmt.Println("1. 一般递归：", num1(n))
fmt.Println("工夫：", time.Now().Sub(now))
fmt.Println("------")
now = time.Now()
fmt.Println("2. 位运算递归：", num2(n))
fmt.Println("工夫：", time.Now().Sub(now))
fmt.Println("------")
now = time.Now()
fmt.Println("3. 我的递归：", num3(n))
fmt.Println("工夫：", time.Now().Sub(now))
}
func num1(n int) int {
if n < 1 {return 0}
record := make([]int, n)
return process1(0, record, n)
}
func process1(i int, record []int, n int) int {
if i == n {return 1}
res := 0
for j := 0; j < n; j++ {if isValid(record, i, j) {record[i] = j
res += process1(i+1, record, n)
}
}
return res
}
func isValid(record []int, i int, j int) bool {
for k := 0; k < i; k++ {if j == record[k] || abs(record[k]-j) == abs(i-k) {return false}
}
return true
}
func abs(a int) int {
if a < 0 {return -a} else {return a}
}
func num2(n int) int {
if n < 1 || n > 32 {return 0}
limit := -1
if n != 32 {limit = (1 << n) - 1
}
return process2(limit, 0, 0, 0)
}
func process2(limit int, colLim int, leftDiaLim int, rightDiaLim int) int {
if colLim == limit {return 1}
pos := limit & (^(colLim | leftDiaLim | rightDiaLim))
mostRightOne := 0
res := 0
for pos != 0 {mostRightOne = pos & (^pos + 1)
pos = pos - mostRightOne
res += process2(limit, colLim|mostRightOne, (leftDiaLim|mostRightOne)<<1,
(rightDiaLim|mostRightOne)>>1)
}
return res
}
func num3(n int) int {rest := make([]int, n)
record := make([]int, n)
for i := 0; i < n; i++ {rest[i] = i
}
ansval := 0
ans := &ansval
process3(record, 0, rest, ans)
return *ans
}
func process3(record []int, recordLen int, rest []int, ans *int) {restLen := len(rest)
if restLen == 0 {
*ans++
return
}
for i := 0; i < restLen; i++ {
isValid := true
for j := 0; j < recordLen; j++ {
// 不须要看同行和同列，只须要思考斜线
if abs(j-recordLen) == abs(record[j]-rest[i]) {
isValid = false
break
}
}
if isValid {record[recordLen] = rest[i]
restCopy := make([]int, restLen)
copy(restCopy, rest)
restCopy = append(restCopy[:i], restCopy[i+1:]...)
process3(record, recordLen+1, restCopy, ans)
}
}
}执行后果如下：
左神 java 代码
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