一道简单的分词程序

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！