福哥答案2021-02-16：
自然智慧即可。
1.普通递归。有代码。
需要判断同列和斜线。
2.位运算递归。有代码。
3.我的递归。有代码。
只需要判断斜线。
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.普通递归：", num1(n))
fmt.Println("时间：", time.Now().Sub(now))
fmt.Println("------")
now = time.Now()
fmt.Println("2.位运算递归：", num2(n))
fmt.Println("时间：", time.Now().Sub(now))
fmt.Println("------")
now = time.Now()
fmt.Println("3.我的递归：", num3(n))
fmt.Println("时间：", 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++ {
//不需要看同行和同列，只需要考虑斜线
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)
}
}
}
执行结果如下：
图片
