Given a string S, find the length of the longest substring T that contains at most two distinct characters.
For example,
Given S = "eceba",
T is "ece" which its length is 3.

给一个字符串，找出最多有两个不同字符的最长子串。还是滑动窗口法，定义一个HashMap或者Array来存储出现过的字符个数，左右指针初始位子为0，右指针向右扫，扫到新字符不同字符数加1，存入HashMap，扫到出现过的字符，HashMap对应字符数量加1。如果不同字符数大于2了，就把左指针位置的字符在HashMap中的数量减1，注意不一定是这一个，如果此时这个字符的数量大于0，说明还有这个字符存在，左指针加1继续右移，扫到的字符以后不会在窗口里了，要在HashMap中减1，如果这个字符减1后为0了，说明这个字符在窗口中没有了，此时不同字符数就可减1，左指针在右移1位指向下一个字符，统计此时窗口长度与最大长度比较，保留最大值。重复上面步骤，直到右指针扫完所有字符。最后返回最大长度。

Java: T: O(n), S: O(1)

public int lengthOfLongestSubstringTwoDistinct(String s) {int max=0;HashMap<Character,Integer> map = new HashMap<Character, Integer>();int start=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);}if(map.size()>2){max = Math.max(max, i-start);while(map.size()>2){char t = s.charAt(start);int count = map.get(t);if(count>1){map.put(t, count-1);}else{map.remove(t);}start++;}}}max = Math.max(max, s.length()-start);return max;
}

Java:

public int lengthOfLongestSubstringTwoDistinct(String s) {Map<Character, Integer> map = new HashMap<>();int start = 0, end = 0;int counter = 0, len = 0;while(end < s.length()){char cur = s.charAt(end);if(!map.containsKey(cur)) map.put(cur, 0);map.put(cur, map.get(cur) + 1);if(map.get(cur) == 1) counter++;//new distinct charwhile(counter > 2){//char c2 = s.charAt(start);map.put(c2, map.get(c2) - 1);if(map.get(c2) == 0) counter--;start++;}len = Math.max(end - start + 1, len);end++;}return len;}

Python:

class Solution:def lengthOfLongestSubstringTwoDistinct(self, s):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 += 1visited[ord(char)] += 1while distinct_count > 2:visited[ord(s[start])] -= 1if visited[ord(s[start])] == 0:distinct_count -= 1start += 1longest = max(longest, i - start + 1)return longest

C++:

class Solution {
public:int lengthOfLongestSubstringTwoDistinct(string s) {int res = 0, left = 0;unordered_map<char, int> m;for (int i = 0; i < s.size(); ++i) {++m[s[i]];while (m.size() > 2) {if (--m[s[left]] == 0) m.erase(s[left]);++left;}res = max(res, i - left + 1);}return res;}
};

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转载于:https://www.cnblogs.com/lightwindy/p/8526298.html

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