[LeetCode] 340. Longest Substring with At Most K Distinct Characters 最多有K个不同字符的最长子串
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Given a string, find the length of the longest substring T that contains at most k distinct characters.
For example, Given s = “eceba”
and k = 2,
T is "ece" which its length is 3.
159. Longest Substring with At Most Two Distinct Characters 的拓展,159题是最多有2个不同字符,这里是最多有k个不同字符。
解题方法一样,只是把比较不同字符个数时的2换成k。
Java:
public class Solution { public int lengthOfLongestSubstringKDistinct(String s, int k) { if (s == null || s.isEmpty() || k < 1) return 0; if (s.length() <= k) return s.length(); int result = k; int pre = 0; Map<Character, Integer> map = new HashMap<>(); for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); map.put(c, map.getOrDefault(c, 0) + 1); while (map.size() > k) { char p = s.charAt(pre++); int count = map.get(p) - 1; if (count == 0) { map.remove(p); } else { map.put(p, count); } } result = Math.max(result, i - pre + 1); } return result; } }
Java:
public class Solution { public int lengthOfLongestSubstringKDistinct(String s, int k) { if (s == null || s.length() == 0 || k <= 0) { return 0; } HashMap<Character, Integer> map = new HashMap<Character, Integer>(); int count = 0; int start = 0; int max = 0; for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (map.containsKey(c)) { map.put(c, map.get(c) + 1); } else { map.put(c, 1); while (map.size() > k) { char rm = s.charAt(start); int tempCount = map.get(rm); if (tempCount > 1) { map.put(rm, tempCount - 1); } else { map.remove(rm); } start++; count--; } } count++; max = Math.max(max, count); } return max; } }
Python:
class Solution(object): def lengthOfLongestSubstringKDistinct(self, s, k): longest, start, distinct_count, visited = 0, 0, 0, [0 for _ in xrange(256)] for i, char in enumerate(s): if visited[ord(char)] == 0: distinct_count += 1 visited[ord(char)] += 1 while distinct_count > k: visited[ord(s[start])] -= 1 if visited[ord(s[start])] == 0: distinct_count -= 1 start += 1 longest = max(longest, i - start + 1) return longest
C++:
class Solution { public: int lengthOfLongestSubstringKDistinct(string s, int k) { int longest = 0, start = 0, distinct_count = 0; array<int, 256> visited = {0}; for (int i = 0; i < s.length(); ++i) { if (visited[s[i]] == 0) { ++distinct_count; } ++visited[s[i]]; while (distinct_count > k) { --visited[s[start]]; if (visited[s[start]] == 0) { --distinct_count; } ++start; } longest = max(longest, i - start + 1); } return longest; } };
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