README_EN.md

1180. Count Substrings with Only One Distinct Letter

中文文档

Description

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 <= s.length <= 1000
• s[i] consists of only lowercase English letters.

Solutions

Python3

class Solution:
    def countLetters(self, s: str) -> int:
        n = len(s)
        i = ans = 0
        while i < n:
            j = i
            while j < n and s[j] == s[i]:
                j += 1
            ans += (1 + j - i) * (j - i) // 2
            i = j
        return ans

class Solution:
    def countLetters(self, s: str) -> int:
        ans = 0
        i, n = 0, len(s)
        while i < n:
            j = i
            cnt = 0
            while j < n and s[j] == s[i]:
                j += 1
                cnt += 1
                ans += cnt
            i = j
        return ans

Java

class Solution {
    public int countLetters(String s) {
        int ans = 0;
        for (int i = 0, n = s.length(); i < n;) {
            int j = i;
            while (j < n && s.charAt(j) == s.charAt(i)) {
                ++j;
            }
            ans += (1 + j - i) * (j - i) / 2;
            i = j;
        }
        return ans;
    }
}

class Solution {
    public int countLetters(String s) {
        int ans = 0;
        int i = 0, n = s.length();
        while (i < n) {
            int j = i;
            int cnt = 0;
            while (j < n && s.charAt(j) == s.charAt(i)) {
                ++j;
                ans += ++cnt;
            }
            i = j;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countLetters(string s) {
        int ans = 0;
        for (int i = 0, n = s.size(); i < n;) {
            int j = i;
            while (j < n && s[j] == s[i]) {
                ++j;
            }
            ans += (1 + j - i) * (j - i) / 2;
            i = j;
        }
        return ans;
    }
};

class Solution {
public:
    int countLetters(string s) {
        int ans = 0;
        int i = 0, n = s.size();
        while (i < n) {
            int j = i;
            int cnt = 0;
            while (j < n && s[j] == s[i]) {
                ++j;
                ans += ++cnt;
            }
            i = j;
        }
        return ans;
    }
};

Go

func countLetters(s string) int {
	ans := 0
	for i, n := 0, len(s); i < n; {
		j := i
		for j < n && s[j] == s[i] {
			j++
		}
		ans += (1 + j - i) * (j - i) / 2
		i = j
	}
	return ans
}

func countLetters(s string) (ans int) {
	i, n := 0, len(s)
	for i < n {
		j := i
		cnt := 0
		for j < n && s[j] == s[i] {
			j++
			cnt++
			ans += cnt
		}
		i = j
	}
	return
}

TypeScript

function countLetters(s: string): number {
    let ans = 0;
    const n = s.length;
    for (let i = 0; i < n; ) {
        let j = i;
        let cnt = 0;
        while (j < n && s[j] === s[i]) {
            ++j;
            ans += ++cnt;
        }
        i = j;
    }
    return ans;
}

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