[LeetCode] 1759. Count Number of Homogenous Substrings

Given a string s, return the number of homogenous substrings of s. Since the answer may be too large, return it modulo 109 + 7.

A string is homogenous if all the characters of the string are the same.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "abbcccaa"
Output: 13
Explanation: The homogenous substrings are listed as below:
"a"   appears 3 times.
"aa"  appears 1 time.
"b"   appears 2 times.
"bb"  appears 1 time.
"c"   appears 3 times.
"cc"  appears 2 times.
"ccc" appears 1 time.
3 + 1 + 2 + 1 + 3 + 2 + 1 = 13.

Example 2:

Input: s = "xy"
Output: 2
Explanation: The homogenous substrings are "x" and "y".

Example 3:

Input: s = "zzzzz"
Output: 15

Constraints:

• 1 <= s.length <= 105
• s consists of lowercase letters.

统计同构子字符串的数目。

给你一个字符串 s ，返回 s 中 同构子字符串 的数目。由于答案可能很大，只需返回对 109 + 7 取余 后的结果。

同构字符串 的定义为：如果一个字符串中的所有字符都相同，那么该字符串就是同构字符串。

子字符串 是字符串中的一个连续字符序列。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/count-number-of-homogenous-substrings
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这是一道数学题。思路是扫描 input 字符串，当遇到连续出现的字符串的时候，统计连续出现的次数 count，同构子字符串的数目 = (1 + count) * count / 2，就是小学数学里的等差数列求和。

时间O(n)

空间O(1)

Java实现

 1 class Solution {
 2     public int countHomogenous(String s) {
 3         final int MOD = (int) Math.pow(10, 9) + 7;
 4         long res = 0;
 5         char pre = s.charAt(0);
 6         int count = 0;
 7         for (int i = 0; i < s.length(); i++) {
 8             char cur = s.charAt(i);
 9             if (cur == pre) {
10                 count++;
11             } else {
12                 res += (long) (count + 1) * count / 2;
13                 count = 1;
14                 pre = cur;
15             }
16         }
17         res += (long) (count + 1) * count / 2;
18         return (int) (res % MOD);
19     }
20 }

LeetCode 题目总结