3. Longest Substring Without Repeating Characters，longestrepeating

3. Longest Substring Without Repeating Characters，longestrepeating

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

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

由题意得，找出有着不重复字符的最长的字符串的长度。题目挺简单的，每次都记录字符串开始的位置start，还有在判断是否重复的时候需要记录重复的下标index，然后遇到重复字符之后的开始位置则为start + index + 1。主函数里是遍历一遍字符串就好了，但是在判断重复字符则也要遍历一遍，所以时间复杂度最坏是在O(n^2)，平均为O(nlogn)，代码如下：

Code(LeetCode运行32ms):

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int Max = 0;
        int length = 0;
        int index = 0;
        int start = 0;
        string temp = "";
        for(int i = 0; i < s.length(); i++) {
            if (!isRepeating(s[i] , temp, index)) {
                length++;
            } else {
                Max = max(length, Max);
                length = temp.length() - index;
                start = start + index + 1;
                temp = s.substr(start, length - 1);
            }
            temp += s[i];
        }
        Max = max(length, Max);
        return Max;
    }
    bool isRepeating(char c, string s, int& index) {
        for (int i = 0; i < s.length(); i++) {
            if (c == s[i]) {
                index = i;
                return true;
            }
        }
        return false;
    }
};

假如改良一下判断重复这一部分的算法，那么可以得到一个O(n)的算法，参考了LeetCode里面大神的做法，使用一个数组来保存某字符上次出现的位置。
Code(LeetCode运行18ms):

int lengthOfLongestSubstring(string s) {
        const int ASCII_MAX = 255;
        int last[ASCII_MAX];
        int start = 0;
        fill(last, last + ASCII_MAX, -1);
        int max_len = 0;
        for (int i = 0; i < s.length(); i++) {
            if (last[s[i]] >= start) {
                max_len = max(max_len, i - start);
                start = last[s[i]] + 1;
            }
            last[s[i]] = i;
        }
        return max((int)s.length() - start, max_len);
    }