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;
}
};
类似题目：
[LeetCode] 3.Longest Substring Without Repeating Characters 最长无重复子串
[LeetCode] 159. Longest Substring with At Most Two Distinct Characters 最多有两个不同字符的最长子串
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