395. Longest Substring with At Least K Repeating Characters

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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 = 3Output:3The longest substring is "aaa", as 'a' is repeated 3 times.Example 2:Input:s = "ababbc", k = 2Output:5The 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。

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