原文:
给定一个字符串 str ，任务是找到最长的前缀，也是给定字符串的后缀。前缀和后缀不应重叠。如果没有这样的前缀，则打印 -1 。
示例:
输入: str = "aabcdaabc" 输出: aabc 字符串“aabc”是最长的 前缀，也是后缀。
输入:str = " aaaa " t3】输出: aa
做法:思路是采用 搜索的预处理算法。在该算法中，我们构建 lps 数组，该数组存储以下值:
lps[i] =最长的前缀 pat[0..i] 也是pat【0】的后缀..i] 。
我们用上面的方法得到长度，然后从前面打印相同数量的字符，这就是我们的答案。
下面是上述方法的实现:
c
// c implementation of the approach
#include
using namespace std;
// returns length of the longest prefix
// which is also suffix and the two do
// not overlap. this function mainly is
// copy of computelpsarray() in kmp algorithm
int lengthlongestprefixsuffix(string s)
{
int n = s.length();
int lps[n];
// lps[0] is always 0
lps[0] = 0;
// length of the previous
// longest prefix suffix
int len = 0;
// loop to calculate lps[i]
// for i = 1 to n - 1
int i = 1;
while (i < n) {
if (s[i] == s[len]) {
len ;
lps[i] = len;
i ;
}
else {
// this is tricky. consider
// the example. aaacaaaa
// and i = 7\. the idea is
// similar to search step.
if (len != 0) {
len = lps[len - 1];
// also, note that we do
// not increment i here
}
// if len = 0
else {
lps[i] = 0;
i ;
}
}
}
int res = lps[n - 1];
// since we are looking for
// non overlapping parts
return (res > n / 2) ? n / 2 : res;
}
// function that returns the prefix
string longestprefixsuffix(string s)
{
// get the length of the longest prefix
int len = lengthlongestprefixsuffix(s);
// stores the prefix
string prefix = "";
// traverse and add characters
for (int i = 0; i < len; i )
prefix = s[i];
// returns the prefix
return prefix;
}
// driver code
int main()
{
string s = "abcab";
string ans = longestprefixsuffix(s);
if (ans == "")
cout << "-1";
else
cout << ans;
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java implementation of the approach
class gfg
{
// returns length of the longest prefix
// which is also suffix and the two do
// not overlap. this function mainly is
// copy of computelpsarray() in kmp algorithm
static int lengthlongestprefixsuffix(string s)
{
int n = s.length();
int lps[] = new int[n];
// lps[0] is always 0
lps[0] = 0;
// length of the previous
// longest prefix suffix
int len = 0;
// loop to calculate lps[i]
// for i = 1 to n - 1
int i = 1;
while (i < n)
{
if (s.charat(i) == s.charat(len))
{
len ;
lps[i] = len;
i ;
}
else
{
// this is tricky. consider
// the example. aaacaaaa
// and i = 7\. the idea is
// similar to search step.
if (len != 0)
{
len = lps[len - 1];
// also, note that we do
// not increment i here
}
// if len = 0
else
{
lps[i] = 0;
i ;
}
}
}
int res = lps[n - 1];
// since we are looking for
// non overlapping parts
return (res > n / 2) ? n / 2 : res;
}
// function that returns the prefix
static string longestprefixsuffix(string s)
{
// get the length of the longest prefix
int len = lengthlongestprefixsuffix(s);
// stores the prefix
string prefix = "";
// traverse and add characters
for (int i = 0; i < len; i )
prefix = s.charat(i);
// returns the prefix
return prefix;
}
// driver code
public static void main(string[] args)
{
string s = "abcab";
string ans = longestprefixsuffix(s);
if (ans == "")
system.out.println("-1");
else
system.out.println(ans);
}
}
python 3
# python 3 implementation of the approach
# returns length of the longest prefix
# which is also suffix and the two do
# not overlap. this function mainly is
# copy of computelpsarray() in kmp algorithm
def lengthlongestprefixsuffix(s):
n = len(s)
lps = [0 for i in range(n)]
# length of the previous
# longest prefix suffix
len1 = 0
# loop to calculate lps[i]
# for i = 1 to n - 1
i = 1
while (i < n):
if (s[i] == s[len1]):
len1 = 1
lps[i] = len1
i = 1
else:
# this is tricky. consider
# the example. aaacaaaa
# and i = 7\. the idea is
# similar to search step.
if (len1 != 0):
len1 = lps[len1 - 1]
# also, note that we do
# not increment i here
# if len = 0
else:
lps[i] = 0
i = 1
res = lps[n - 1]
# since we are looking for
# non overlapping parts
if (res > int(n / 2)):
return int(n / 2)
else:
return res
# function that returns the prefix
def longestprefixsuffix(s):
# get the length of the longest prefix
len1 = lengthlongestprefixsuffix(s)
# stores the prefix
prefix = ""
# traverse and add characters
for i in range(len1):
prefix = s[i]
# returns the prefix
return prefix
# driver code
if __name__ == '__main__':
s = "abcab"
ans = longestprefixsuffix(s)
if (ans == ""):
print("-1")
else:
print(ans)
# this code is contributed by
# surendra_gangwar
c
// c# implementation of the approach
using system;
class gfg
{
// returns length of the longest prefix
// which is also suffix and the two do
// not overlap. this function mainly is
// copy of computelpsarray() in kmp algorithm
static int lengthlongestprefixsuffix(string s)
{
int n = s.length;
int []lps = new int[n];
// lps[0] is always 0
lps[0] = 0;
// length of the previous
// longest prefix suffix
int len = 0;
// loop to calculate lps[i]
// for i = 1 to n - 1
int i = 1;
while (i < n)
{
if (s[i] == s[len])
{
len ;
lps[i] = len;
i ;
}
else
{
// this is tricky. consider
// the example. aaacaaaa
// and i = 7\. the idea is
// similar to search step.
if (len != 0)
{
len = lps[len - 1];
// also, note that we do
// not increment i here
}
// if len = 0
else
{
lps[i] = 0;
i ;
}
}
}
int res = lps[n - 1];
// since we are looking for
// non overlapping parts
return (res > n / 2) ? n / 2 : res;
}
// function that returns the prefix
static string longestprefixsuffix(string s)
{
// get the length of the longest prefix
int len = lengthlongestprefixsuffix(s);
// stores the prefix
string prefix = "";
// traverse and add characters
for (int i = 0; i < len; i )
prefix = s[i];
// returns the prefix
return prefix;
}
// driver code
public static void main()
{
string s = "abcab";
string ans = longestprefixsuffix(s);
if (ans == "")
console.writeline("-1");
else
console.writeline(ans);
}
}
// this code is contributed by ryuga
java 描述语言
output:
ab
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