**题目：**
> 给出字符串dogandcat及字典{"do", "d", "gand", "g", "and", "dog", "cat", "c", "at", "in", "play", "i", "nplay"}，
> 找出组成字符串的所以可能的单词组合。
之前用haskell写过一个实现，但接触golang之后，发现可以写成更简明的版本，代码如下：
```
package main
import (
"fmt"
)
var word = "dogandcat"
var dicts = []string{"do", "d", "gand", "g", "and", "dog", "cat", "c", "at", "in", "play", "i", "nplay"}
var results = [][]string{}
var cpyCounter = 0
var stepCounter = 0
func IsContainsInDicts(str string) bool {
for _, v := range dicts {
if v == str {
return true
}
}
return false
}
func breaker(matched []string, pos int) {
if pos >= len(word) {
results = append(results, matched)
return
}
match := []byte{}
for i := pos; i < len(word); i++ {
match = append(match, word[i])
aword := string(match)
stepCounter += 1
if IsContainsInDicts(aword) {
cpyMatched := make([]string, len(matched))
copy(cpyMatched, matched)
cpyMatched = append(cpyMatched, aword)
cpyCounter += 1
breaker(cpyMatched, i+1)
}
}
}
func main() {
fmt.Println(word)
fmt.Println(dicts)
breaker([]string{}, 0)
fmt.Println()
fmt.Println(results)
fmt.Println()
fmt.Println("step times: ", stepCounter)
fmt.Println("cpy times: ", cpyCounter)
}
```
以上代码还有点性能上的小问题：
- 1）如果字典很大，IsContainsInDicts的效率就很低；
- 2）如果字符串很大，result很耗内存；
如果有更好的方法，欢迎指教，谢谢！
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