Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"Output: 3 Explanation: The answer is "abc", which the length is 3.

Solution:

int lengthOfLongestSubstring(string s) {    int max = 0, temp = 0, now = 0;    unordered_map
     
       c2i;    for (int  i = 0; i < s.length(); i++){        if(c2i.count(s[i]) == 0){            temp++;            c2i[s[i]] = i;        }else{            int t = c2i[s[i]];            c2i[s[i]] = i;            for(int j = t-1; j >= now; j--){                c2i.erase(s[j]);            }            if(temp > max){                max = temp;            }            temp -= (t - now);            now = t + 1;        }    }    max = max > temp ? max : temp;    return max;}
     

PS.

今天的问题只要在于对于map的操作不熟悉，应用的几个操作分别是：

count(KEY)：统计某个值出现的次数（0或1因为map不允许出现重复元素）；

erase(KEY)：抹除某个值的元素。

除此之外查看C++文档：

http://www.cplusplus.com/reference/unordered_map/unordered_map/

PPS.

看到一种更好的写法（JAVA）

public int lengthOfLongestSubstring(String s) {    int n = s.length();    Set
     
       set = new HashSet<>();    int ans = 0, i = 0, j = 0;    while (i < n && j < n) {        // try to extend the range [i, j]        if (!set.contains(s.charAt(j))){            set.add(s.charAt(j++));            ans = Math.max(ans, j - i);        }        else {            set.remove(s.charAt(i++));        }    }    return ans;}