//思路：参考Mood 清华大学《算法与分析》汪小林 第三周：分支策略（2） 3.4芯片测试
//时间复杂度O(n)
func main() {
//true：好芯片，false:坏芯片
//chips := []bool{true,false,true}
chips := []chip{
{
good: false,
num: 0,
},{
good: false,
num: 1,
},{
good: false,
num: 2,
},{
good: false,
num: 3,
},{
good: true,
num: 4,
},{
good: true,
num: 5,
},{
good: true,
num: 6,
},{
good: true,
num: 7,
},{
good: true,
num: 8,
}}
fmt.Println(getGoodChip(chips))
}
type chip struct {
good bool
num int
}
//return good chip index
func getGoodChip(chips []chip) int{
if len(chips) == 3 {
if judge(chips[0].good,chips[1].good){
return chips[0].num
}else {
return chips[2].num
}
}
if len(chips) <=2 {
return chips[0].num
}
if len(chips) %2 == 0{
return getGoodChip(evenChips(chips))
}
return getGoodChip(oddChips(chips))
}
func evenChips(theChips []chip) []chip {
newChips := []chip{}
for i:=0; i<len(theChips) ;i+=2 {
if judge(theChips[i].good, theChips[i+1].good){
newChips = append(newChips, theChips[i])
}
}
return newChips
}
func oddChips(theChips []chip) []chip {
candChip := theChips[0]
n := len(theChips)
goodNum := 0
for i:=1;i<n ; i++ {
if judge(candChip.good,theChips[i].good){
goodNum++
}
}
if goodNum >= (n-1)/2 {//find good chip ,return it.End.
return []chip{candChip}
}
//else candChip is a bad chip ,abandon it
return theChips[1:]
}
//1：芯片x判断芯片y为好芯片，否则为0
func judge(a,b bool)bool{
if a && b {
return true
}
if !a && !b {
aRand := rand.Intn(2)
bRand := rand.Intn(2)
if aRand ==1 && bRand ==1{
return true
}else{
return false
}
}
return false
}