1180. Count Substrings with Only One Distinct Letter 统计只含单一字母的子串

• 作者： 负雪明烛
• id： fuxuemingzhu
• 个人博客：http://fuxuemingzhu.cn/

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@TOC

题目地址：https://leetcode-cn.com/problems/count-substrings-with-only-one-distinct-letter/

题目描述

Given a string S, return the number of substrings that have only one distinct letter.

Example 1:

Input: S = "aaaba"
Output: 8
Explanation: The substrings with one distinct letter are "aaa", "aa", "a", "b".
"aaa" occurs 1 time.
"aa" occurs 2 times.
"a" occurs 4 times.
"b" occurs 1 time.
So the answer is 1 + 2 + 4 + 1 = 8.

Example 2:

Input: S = "aaaaaaaaaa"
Output: 55

Constraints:

1. 1 <= S.length <= 1000
2. S[i] consists of only lowercase English letters.

题目大意

给你一个字符串 S，返回只含 单一字母 的子串个数。

解题方法

组合数

从一段长度为n的单一字母字符串中，分别选择出长度为1,2,3,...,n的子串，共有n * (n + 1) / 2个结果。

因此，遍历字符串，分别统计出单一字母字符串的长度，累加所有结果即可。

C++代码如下：

class Solution {
public:
    int countLetters(string S) {
        if (S.empty()) return 0;
        int count = 1;
        int res = 0;
        char cur = S[0];
        cout << endl;
        for (int i = 1; i <= S.size(); ++i) {
            if (i == S.size() || S[i] != cur) {
                res += count * (count + 1) / 2;
                cur = S[i];
                count = 1;
            } else {
                count ++;
            }
            
        }
        return res;
    }
};

日期

2019 年 9 月 18 日 —— 今日又是九一八