2730. 找到最长的半重复子字符串

class Solution:
    def longestSemiRepetitiveSubstring(self, s: str) -> int:
        n = len(s)
        ans = cnt = j = 0
        for i in range(n):
            if i and s[i] == s[i - 1]:
                cnt += 1
            while cnt > 1:
                if s[j] == s[j + 1]:
                    cnt -= 1
                j += 1
            ans = max(ans, i - j + 1)
        return ans

class Solution {
    public int longestSemiRepetitiveSubstring(String s) {
        int n = s.length();
        int ans = 0;
        for (int i = 0, j = 0, cnt = 0; i < n; ++i) {
            if (i > 0 && s.charAt(i) == s.charAt(i - 1)) {
                ++cnt;
            }
            while (cnt > 1) {
                if (s.charAt(j) == s.charAt(j + 1)) {
                    --cnt;
                }
                ++j;
            }
            ans = Math.max(ans, i - j + 1);
        }
        return ans;
    }
}

class Solution {
public:
    int longestSemiRepetitiveSubstring(string s) {
        int n = s.size();
        int ans = 0;
        for (int i = 0, j = 0, cnt = 0; i < n; ++i) {
            if (i && s[i] == s[i - 1]) {
                ++cnt;
            }
            while (cnt > 1) {
                if (s[j] == s[j + 1]) {
                    --cnt;
                }
                ++j;
            }
            ans = max(ans, i - j + 1);
        }
        return ans;
    }
};

func longestSemiRepetitiveSubstring(s string) (ans int) {
	n := len(s)
	for i, j, cnt := 0, 0, 0; i < n; i++ {
		if i > 0 && s[i] == s[i-1] {
			cnt++
		}
		for cnt > 1 {
			if s[j] == s[j+1] {
				cnt--
			}
			j++
		}
		ans = max(ans, i-j+1)
	}
	return
}

function longestSemiRepetitiveSubstring(s: string): number {
    const n = s.length;
    let ans = 0;
    for (let i = 0, j = 0, cnt = 0; i < n; ++i) {
        if (i > 0 && s[i] === s[i - 1]) {
            ++cnt;
        }
        while (cnt > 1) {
            if (s[j] === s[j + 1]) {
                --cnt;
            }
            ++j;
        }
        ans = Math.max(ans, i - j + 1);
    }
    return ans;
}