洗牌算法：产生一个随机序列
```go
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
fmt.Println(Shuffle([]int{10,20,30,40}))
}
func Shuffle(vals []int) []int {
newVals := make([]int,0,len(vals))
r := rand.New(rand.NewSource(time.Now().Unix()))
for _,i := range r.Perm(len(vals)) {
newVals = append(newVals, vals[i])
}
return newVals
}
```
关键步骤：r.Perm()，解释：
```go
// Perm returns, as a slice of n ints, a pseudo-random permutation of the integers [0,n).
func (r *Rand) Perm(n int) []int {
m := make([]int, n)
// In the following loop, the iteration when i=0 always swaps m[0] with m[0].
// A change to remove this useless iteration is to assign 1 to i in the init
// statement. But Perm also effects r. Making this change will affect
// the final state of r. So this change can't be made for compatibility
// reasons for Go 1.
for i := 0; i < n; i++ {
j := r.Intn(i + 1)
m[i] = m[j]
m[j] = i
}
return m
}
```
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