Leetcode_Longest Substring Without Repeating Characters

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leetcode倒数第三题，地址：http://oj.leetcode.com/problems/longest-substring-without-repeating-characters/

思路：用deque和map来模拟，从后插入deque，同时map存一份表示此char已存入，当发现当前char已存入deque的时候，从deque的front开始pop，知道把与当前char相同的char 给pop出去，同时map也做维护，当q的size最大时，也就是要返回的值。

形象上就是两个指针一前一后在移动，有char进来并不与两指针之间的char重复的时候，后指针移动(往后)，否则前指针移动(往后)到重复char的后一位置。map维护快速查询。

参考代码：

class Solution {public:    int lengthOfLongestSubstring(string s) {        int maxlen = 0, len = 0;        deque<char> q;        map<char, bool> resmap;        for(int i = 0; i < s.length(); ++i)        {            auto it = resmap.find(s[i]);            if(it == resmap.end())            {                ++len;                q.push_back(s[i]);                maxlen = maxlen > len ? maxlen : len;                resmap.insert(pair<char,bool>(s[i], true));            }            else            {                while(!q.empty() && q.front() != s[i])                {                    auto deleteIt = resmap.find(q.front());                    resmap.erase(deleteIt);                    q.pop_front();                }                q.pop_front();                q.push_back(s[i]);                len = q.size();            }        }        return maxlen;    }};

SECOND TRIAL, 204MS

class Solution {public:    int lengthOfLongestSubstring(string s) {        if(s.empty())            return 0;        unordered_map<char, int>mp;        unordered_map<char, int>::iterator it;        int i = 0, mlen = 0;        for(int k = 0; k<s.length(); ++k)        {            it = mp.find(s[k]);            if(it == mp.end())                mp[s[k]]=k;            else            {                while(s[i]!=s[k])                {                    it = mp.find(s[i]);                    mp.erase(it);                    ++i;                }                ++i;            }            mlen = max(mlen, k-i+1);        }        return mlen;    }};

PYTHON VERSION, 392MS

class Solution:    # @return an integer    def lengthOfLongestSubstring(self, s):        if not s:            return 0        d = {}        maxlen = i = 0        for j in range(0, len(s)):            if not d.has_key(s[j]):                d[s[j]] = j            else:                while s[i] != s[j]:                    del d[s[i]]                    i+=1                i+=1            maxlen = max(maxlen, j-i+1)        return maxlen

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