LeetCode-3-Longest Substring Without Repeating Characters

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.

时间复杂度O(n^2)

#include <iostream>
#include <string>
using namespace std;


class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        size_t found;
        int max_len = s.empty() ? 0 : 1;
        size_t prev = 0, next = 1;
        int len = 1;
        while (next < s.size()) {
            found = find(s.at(next), s, prev, next);
            if (found == string::npos) {
                ++len;
                ++next;
            } else {
                if (len > max_len) {
                    max_len = len;
                }
                prev = found + 1;
                len = next - prev + 1;
                ++next;
            }


        }
        // 防止最后一个窗口没有将较大的len赋值给max_len,同时考虑s为空的情形
        if (len > max_len && !s.empty()) {
            max_len = len;
        }
        return max_len;
    }


private:
    size_t find(char ch, string s, size_t start_pos, size_t end_pos) {
        size_t pos = start_pos;
        while (pos < end_pos) {
            if (s.at(pos) == ch) {
                return pos;
            }
            pos++;
        }
        return string::npos;
    }
};


int main() {
    string str1 = "abcabcbb";
    string str2 = "bbbbb";
    string str3 = "pwwkew";
    string str4 = "au";
    string str5 = "a";
    string str6;
    Solution s;
    cout << s.lengthOfLongestSubstring(str1) << endl
         << s.lengthOfLongestSubstring(str2) << endl
         << s.lengthOfLongestSubstring(str3) << endl
         << s.lengthOfLongestSubstring(str4) << endl
         << s.lengthOfLongestSubstring(str5) << endl
         << s.lengthOfLongestSubstring(str6) << endl;


return 0;
}