给定一个数n如23121;给定一组数字a如[2 4 9]求由a中元素组成的小于n的最大数

给你一个n和一个数字的数组，比如n=23121，数组A={2,4,9},当然保证数组A中都是个位数，并且没有重复的，没说数组A有序，我这默认有序，结果得到一个数22999，就是数组A拼出来的一个刚好比n小的数。

 

package main

import (
    "fmt"
    "strings"
)
func main() {
    num := 45543
    dstArr := []string{"4","9","2"}
    numList := strings.Split(fmt.Sprintf("%d",num),"")
    ret := solution2(numList,dstArr,false)
    fmt.Println(ret)
}

// 23121
// [2,4,9]
func solution2(num []string, arr []string, pre bool) []string {
    if len(num) == 1 && !pre {
        ret := lastNum(num[0],arr)
        if len(ret) == 0 {
            return []string{}
        }
        return []string{ret}
    } else if len(num) == 1 {
        return []string{maxNum(arr)}
    }
    def := make([]string, 0)
    curNun := num[0]
    if pre {
        def = append(def,maxNum(arr))
        def = append(def,solution2(num[1:], arr, pre)...)
        return def
    }
    e := equNum(curNun,arr)
    ret1 := make([]string, 0)
    if len(e) != 0 {
        ret := solution2(num[1:], arr, pre)
        if len(ret) > 0 {
            ret1 = append(ret1,e)
            ret1 = append(ret1,ret...)
        }
    }
    m := lastNum(curNun,arr)
    ret2 := make([]string, 0)
    if len(m) != 0 {
        ret := solution2(num[1:], arr, true)
        if len(ret) > 0 {
            ret2 = append(ret2,m)
            ret2 = append(ret2,ret...)
        }
    }
    ret3 := solution2(num[1:], arr, true)
    if len(ret1) > 0 {
        return ret1
    } else if len(ret2) > 0 {
        return ret2
    }
    return ret3

}

func maxNum(arr []string) string{
    var p string
    for _, item :=range arr {
        if p < item {
            p = item
        }
    }
    return p
}

func lastNum(n string, arr []string) string{
    var last string
    for _, item :=range arr {
        if item < n && last < item {
            last = item
        }
    }
    return last
}

func equNum(n string, arr []string) string{
    for _, item :=range arr {
        if item == n  {
            return item
        }
    }
    return ""
}