题目描述:
求字符串中出现次数最多的子串的出现次数。
例如字符串abcbcbcabc,出现次数最多的子串是bc,出现次数为4
思路:利用后缀数组:
abcbcbcabc 第0个
bcbcbcabc 第1个
cbcbcabc 第2个
bcbcabc 第3个
cbcabc 第4个
bcabc 第5个
cabc 第6个
abc 第7个
bc 第8个
c 第9个
过程:先从第0个数组取出a,然后和第1个数组的b比较,发现不相等,然后取ab,abc,abcb....这一趟取完后,又从第1个后缀数组开始取,取b,bc,bcb,bcbc...
#include<iostream>
#include<string>
#include<vector>
using namespace std;
pair<int, string> fun(const string &str)
{
vector<string> substrs;
int len = str.length();
string substring;
int maxcount(0);
//后缀数组
cout << "the string is:" << str << endl;
cout << "the substrings are as follows:" << endl;
for (int i = 0; i < len; ++i)
{
substrs.push_back(str.substr(i));
cout << substrs[i] << endl;
}
cout << "--------------the answer------------" << endl;
for (int i = 0; i < len; ++i)
{
for (int j = 1; j <= len; j++) {
int count = 1;
int sublen = j;
for (int k = i + 1; k < len; k++) {
if (substrs[k].length() < sublen) {
break;
}
//cout << substrs[i].substr(0, sublen) << endl;
//cout << substrs[k].substr(0, sublen) << endl;
string str1 = substrs[i].substr(0, sublen);
string str2 = substrs[k].substr(0, sublen);
//cout << "比较结果:" << str1.compare(str2) << endl;
//cout << "i = " << i << " sublen = " << j << " k = " << k << endl;
if (str1.compare(str2)==0)
{
++count;
}
//cout << "count = " << count << endl;
}
if (count > maxcount||(count == maxcount && sublen > substring.length()))
{
maxcount = count;
substring = substrs[i].substr(0, sublen);
}
}
}
return make_pair(maxcount, substring);
}
int main()
{
string str = "ababcababcabcab";
auto res = fun(str);
cout << "the max count is:" << res.first << endl;
cout << "the matched substring is:" << res.second << endl;
return 0;
}
输出:
the string is:ababcababcabcab
the substrings are as follows:
ababcababcabcab
babcababcabcab
abcababcabcab
bcababcabcab
cababcabcab
ababcabcab
babcabcab
abcabcab
bcabcab
cabcab
abcab
bcab
cab
ab
b
--------------the answer------------
the max count is:6
the matched substring is:ab
参考资料:
[1] 求一个字符串中连续出现次数最多的子串 https://blog.csdn.net/qq_22080999/article/details/81143555
[2] 【算法刷题】一个字符串中连续出现次数最多的子串 https://blog.csdn.net/Neo_dot/article/details/80559744