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1698. 字符串的不同子字符串个数

English Version

题目描述

给定一个字符串 s，返回 s 的不同子字符串的个数。

字符串的 子字符串 是由原字符串删除开头若干个字符（可能是 0 个）并删除结尾若干个字符（可能是 0 个）形成的字符串。

 

示例 1：

输入：s = "aabbaba"
输出：21
解释：不同子字符串的集合是 ["a","b","aa","bb","ab","ba","aab","abb","bab","bba","aba","aabb","abba","bbab","baba","aabba","abbab","bbaba","aabbab","abbaba","aabbaba"]

示例 2：

输入：s = "abcdefg"
输出：28

 

提示：

• 1 <= s.length <= 500
• s 由小写英文字母组成。

进阶：你可以以 O(n) 时间复杂度解决此问题吗？

解法

Python3

class Solution:
    def countDistinct(self, s: str) -> int:
        n = len(s)
        return len({s[i:j] for i in range(n) for j in range(i + 1, n + 1)})

Java

class Solution {
    public int countDistinct(String s) {
        Set<String> ss = new HashSet<>();
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j <= n; ++j) {
                ss.add(s.substring(i, j));
            }
        }
        return ss.size();
    }
}

C++

class Solution {
public:
    int countDistinct(string s) {
        unordered_set<string_view> ss;
        int n = s.size();
        string_view t, v = s;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j <= n; ++j) {
                t = v.substr(i, j - i);
                ss.insert(t);
            }
        }
        return ss.size();
    }
};

Go

func countDistinct(s string) int {
	ss := map[string]bool{}
	for i := range s {
		for j := i + 1; j <= len(s); j++ {
			ss[s[i:j]] = true
		}
	}
	return len(ss)
}

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