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.
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 {
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;
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;
}
};
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
}