给定一个字符串，在 Golang 中打印不重复字符的最长子字符串。例如，“ABDEFGABEF”没有重复字符的最长子串是“BDEFGA”。
例子：
Input : GEEKSFORGEEKS
Output :
Substring: EKSFORG
Length: 7
Input : ABDEFGABEF
Output :
Substring: BDEFGA
Length: 6
这个想法是遍历字符串，对于每个已经访问过的字符，将其最后一次出现在整数数组pos中（我们将根据字符串字符的 ASCII 值更新数组中的索引）。变量st存储当前子串的起始点， maxlen存储最大长度子串的长度，start存储最大长度子串的起始索引。
在遍历字符串，通过检查数组pos来检查当前字符是否已经存在于字符串。如果不存在，则将当前字符的索引存储在数组中，并将 value 作为当前索引。如果它已经存在于数组中，这意味着当前字符可以在当前子字符串中重复。为此，请检查上一次出现的字符是在当前子字符串的起点 st 之前还是之后。如果它在st之前，则只更新数组中的值。如果在 st 之后，则求当前子串currlen的长度为i-st ，其中i 是当前索引。
将currlen与maxlen进行比较。如果maxlen小于 currlen，则将 maxlen 更新为currlen并从 st 开始。完全遍历字符串，需要的最长不重复字符的子串是从s[start] 到 s[start+maxlen-1] 。
// Golang program to find the length of the
// longest substring without repeating
// characters
package main
import "fmt"
func longest_substring(s string, n int) string {
var i int
// starting point of
// current substring.
st := 0
// length of current substring.
currlen := 0
// maximum length substring
// without repeating characters
maxlen := 0
// starting index of
// maximum length substring.
start := 0
// this array works as the hash table
// -1 indicates that element is not
// present before else any value
// indicates its previous index
pos := make([]int, 125)
for i = 0; i < 125; i++ {
pos[i] = -1
}
// storing the index
// of first character
pos[s[0]] = 0
for i = 1; i < n; i++ {
// If this character is not present in array,
// then this is first occurrence of this
// character, store this in array.
if pos[s[i]] == -1 {
pos[s[i]] = i
} else {
// If this character is present in hash then
// this character has previous occurrence,
// check if that occurrence is before or after
// starting point of current substring.
if pos[s[i]] >= st {
// find length of current substring and
// update maxlen and start accordingly.
currlen = i - st
if maxlen < currlen {
maxlen = currlen
start = st
}
// Next substring will start after the last
// occurrence of current character to avoid
// its repetition.
st = pos[s[i]] + 1
}
// Update last occurrence of
// current character.
pos[s[i]] = i
}
}
// Compare length of last substring
// with maxlen and update maxlen
// and start accordingly.
if maxlen < i-st {
maxlen = i - st
start = st
}
// the required string
ans := ""
// extracting the string from
// [start] to [start+maxlen]
for i = start; i < start+maxlen; i++ {
ans += string(s[i])
}
return ans
}
func main() {
var s string = "GEEKSFORGEEKS"
var n int = len(s)
// calling the function to
// get the required answer.
var newString = longest_substring(s, n)
// finding the length of the string
var length int = len(newString)
fmt.Println("Longest substring is: ", newString)
fmt.Println("Length of the string: ", length)
}
输出：
Longest substring is: EKSFORG
Length of the string: 7
注意：除了使用数组，我们还可以使用字符串到int的映射。