395. Longest Substring with At Least K Repeating Characters，longestrepeating

395. Longest Substring with At Least K Repeating Characters，longestrepeating

Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times.

Example 1:

Input:
s = "aaabb", k = 3

Output:
3

The longest substring is "aaa", as 'a' is repeated 3 times.
Example 2:

Input:
s = "ababbc", k = 2

Output:
5

The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.

这题如果逐一查找的话，操作起来不太方便，这题最方便的做法就是递归了。建立一个数组来统计每个字母出现的次数，对于出现次数少于K次的，我们把它们当做间隔，从而把原数列分为几段，要找的满足重复的最长的子序列一定在这几段其中。对于分开的每段，又是一个同样性质的子问题。如果一段中没有小于K的间隔，那么直接返回这一段的长度，更新最大长度。

int longestSubstring(string s, int k) {
    return helper(s, k, 0, s.size() - 1);
}
int helper(string s, int k, int left, int right) {
    int len = right - left + 1;
    if (len <= 0) return 0;
    int i, j;
    int maxlen = 0;
    vector<int> count(26, 0);
    for (i = left; i <= right; i++) {
        count[s[i] - 'a']++;
    }
    for (i = left, j = left; i <= right; i++) {
        if (count[s[i] - 'a'] < k) {
            maxlen = max(maxlen, helper(s, k, j, i - 1));
            j = i + 1;
        }

    }
    if (j == left) return len;
    else return max(maxlen, helper(s, k, j, i - 1));
}

注意一下，最后一句的处理：

else return max(maxlen, helper(s, k, j, i - 1));

如果在原数列中出现了间隔，那么最后一个间隔到right之间这一段是没有参与计数，这个很容易被忽视。leetcode有测试用例特意测了这一点。