我已經(jīng)實(shí)現(xiàn)了最長(zhǎng)公共子序列算法并獲得了最長(zhǎng)的正確答案，但無(wú)法弄清楚打印出最長(zhǎng)公共子序列的組成部分的方法。也就是說(shuō)，我成功獲得了最長(zhǎng)公共子序列數(shù)組的長(zhǎng)度，但我想打印出最長(zhǎng)的子序列。此代碼的游樂(lè)場(chǎng)在這里http://play.golang.org/p/0sKb_OARnf/*X = BDCABAY = ABCBDAB => Longest Comman Subsequence is B C BDynamic Programming method : O ( n )*/package mainimport "fmt"func Max(more ...int) int {  max_num := more[0]  for _, elem := range more {    if max_num < elem {      max_num = elem    }  }  return max_num}func Longest(str1, str2 string) int {  len1 := len(str1)  len2 := len(str2)  //in C++,  //int tab[m + 1][n + 1];  //tab := make([][100]int, len1+1)  tab := make([][]int, len1+1)  for i := range tab {    tab[i] = make([]int, len2+1)  }  i, j := 0, 0  for i = 0; i <= len1; i++ {    for j = 0; j <= len2; j++ {      if i == 0 || j == 0 {        tab[i][j] = 0      } else if str1[i-1] == str2[j-1] {        tab[i][j] = tab[i-1][j-1] + 1        if i < len1 {          fmt.Printf("%c", str1[i])        }      } else {        tab[i][j] = Max(tab[i-1][j], tab[i][j-1])      }    }  }  fmt.Println()  return tab[len1][len2]}func main() {  str1 := "AGGTABTABTABTAB"  str2 := "GXTXAYBTABTABTAB"  fmt.Println(Longest(str1, str2))  //Actual Longest Common Subsequence: GTABTABTABTAB  //GGGGGTAAAABBBBTTTTAAAABBBBTTTTAAAABBBBTTTTAAAABBBB  //13  str3 := "AGGTABGHSRCBYJSVDWFVDVSBCBVDWFDWVV"  str4 := "GXTXAYBRGDVCBDVCCXVXCWQRVCBDJXCVQSQQ"  fmt.Println(Longest(str3, str4))  //Actual Longest Common Subsequence: ?  //GGGTTABGGGHHRCCBBBBBBYYYJSVDDDDDWWWFDDDDDVVVSSSSSBCCCBBBBBBVVVDDDDDWWWFWWWVVVVVV  //14}當(dāng)我嘗試在選項(xiàng)卡更新時(shí)打印出子序列時(shí)，結(jié)果是重復(fù)的。我想為 str1 和 str2 打印出類似“GTABTABTABTAB”的內(nèi)容