LeetCode题解3：Longest Substring Without Repeating Characters

Longest Substring Without Repeating Characters问题：求给定字符串的最大无重复字符的子串的长度。

难度：中等

思路：用2个变量（start, end）分别保存子串的起点和终点。end自增，直到遇到重复字符为止；从重复字符出现的位置之后1位重新开始扫描。

陷阱：重新开始扫描的位置；循环结束条件

代码：

class Solution:
    # @param {string} s
    # @return {integer}
    def lengthOfLongestSubstring(self, s):
        len_max = 0
        start = 0
        sub_string = ''
        for end in xrange(len(s)):
            if s[end] not in sub_string:
                sub_string += s[end]
            else:
                len_max = max(len_max, len(sub_string))
                while s[start] != s[end]:
                    start += 1
                start += 1
                sub_string = s[start : end + 1]
        return max(len_max, len(sub_string))

以上解法可求出最大无重复字符的子串本身，由于题目只要求求出最大无重复字符的子串的长度，因此可用Python集合（set）类型略作优化如下：

class Solution:
    # @param {string} s
    # @return {integer}
    def lengthOfLongestSubstring(self, s):
        length = 0
        char_set = set()
        start = end = 0
        while start < len(s) and end < len(s):
            if s[end] in char_set:
                length = max(length, end - start)
                char_set.remove(s[start])
                start += 1
            else:
                char_set.add(s[end])
                end += 1
        return max(length, end - start)

另外，用Python字典（dict）进行上述优化也是可行的，该字典记录了字符到下标的反向索引，代码如下：

class Solution:
    # @param {string} s
    # @return {integer}
    def lengthOfLongestSubstring(self, s):
        start = len_max = 0
        char_dict = {}
        for i in range(len(s)):
            if s[i] in char_dict and start <= char_dict[s[i]]:
                start = char_dict[s[i]] + 1
            else:
                len_max = max(len_max, i - start + 1)
            char_dict[s[i]] = i
        return len_max

最后，使用enumerate和Python字典的get方法可使代码更简洁：

class Solution:
    # @param {string} s
    # @return {integer}
    def lengthOfLongestSubstring(self, s):
        max_length, start, char_dict = 0, 0, {}
        for idx, char in enumerate(s, 1):
            if char_dict.get(char, -1) >= start:
                start = char_dict[char]
            char_dict[char] = idx
            max_length = max(max_length, idx - start)
        return max_length