LeetCode 159. Longest Substring with At Most Two Distinct Characters --Java,C++,Python解法

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LeetCode题解专栏：LeetCode题解
我做的所有的LeetCode的题目都放在这个专栏里，大部分题目Java和Python的解法都有。

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题目地址：Longest Substring with At Most Two Distinct Characters - LeetCode

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Given a string s , find the length of the longest substring t that contains at most 2 distinct characters.

Example 1:

Input: "eceba"
Output: 3
Explanation: t is "ece" which its length is 3.

Example 2:

Input: "ccaabbb"
Output: 5
Explanation: t is "aabbb" which its length is 5.

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这道题目是hard难度的，题目的意思很简单，就是找到具有最多两个不同字符的最长子串的长度。

看到题目会感觉O(n)的时间复杂度应该能做出这道题目。

在写了大约20分钟代码后，Python解法出来了：

class Solution:
    def lengthOfLongestSubstringTwoDistinct(self, s: str) -> int:
        count = 0
        p = {}
        l = {}
        for now in range(0, len(s)):
            i = s[now]
            p[i] = now
            if len(l) < 2 and i not in l:
                l[i] = 1
            elif i in l:
                l[i] += 1
            else:
                if count < sum(l.values()):
                    count = sum(l.values())
                for j in l.keys():
                    if j != s[now-1]:
                        l.pop(j)
                        l[list(l.keys())[0]] = now-p[j]-1
                        break
                l[i] = 1
        if l != {}:
            if count < sum(l.values()):
                count = sum(l.values())
        return count

我的解法稍微有点麻烦，但易懂，最快解法如下：

class Solution:
    def lengthOfLongestSubstringTwoDistinct(self, s: str) -> int:
        i, l, k = -1, 0, 0
        for j in range(1, len(s)):
            if s[j] != s[j-1]:
                if i > -1 and s[i] != s[j]:
                    l, k = i+1, max(k, j-l)
                i = j-1
        return max(k, len(s)-l)

核心思路跟我的做法一样。

区别在于它使用i和j来跟踪两个字符的最后一个索引。i跟踪第一个字符，j跟踪第二个字符。但我仍然喜欢hashmap解决方案，因为我们可以轻松地将该解决方案扩展到K个不同的字符。

C++解法如下：

class Solution
{
public:
    int lengthOfLongestSubstringTwoDistinct(string s)
    {
        int i = 0, j = -1;
        int maxLen = 0;
        for (int k = 1; k < s.size(); k++)
        {
            if (s[k] == s[k - 1])
                continue;
            if (j > -1 && s[k] != s[j])
            {
                maxLen = max(maxLen, k - i);
                i = j + 1;
            }
            j = k - 1;
        }
        return maxLen > (s.size() - i) ? maxLen : s.size() - i;
    }
};

其他人的Java解法如下：

class Solution {
    public int lengthOfLongestSubstringTwoDistinct(String str) {
        int i = 0, j = -1;
        int maxLen = 0;
        char[] s = str.toCharArray();
        for (int k = 1; k < s.length; k++) {
            if (s[k] == s[k - 1])
                continue;
            if (j > -1 && s[k] != s[j]) {
                maxLen = Math.max(maxLen, k - i);
                i = j + 1;
            }
            j = k - 1;
        }
        return maxLen > (s.length - i) ? maxLen : s.length - i;
    }
}