3. Longest Substring Without Repeating Characters

1. 问题：

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.

2方法:

     方法1. 最常见的冒泡：  Me

class Solution {public:    int lengthOfLongestSubstring(string s)     {        string  tmp_result;        int tmp_maxlen = 0;            int len = s.length();        int j = 0;        int i = 0;        int maxlen = 0;        int flag_break = 0;        for (i=0; i<len; i++)        {            for (j=i; j<len; j++)            {            //  cout << "i:" << i << ",s[j]"<< s.at(j) << endl;                if (tmp_result.find(s.at(j)) == string::npos)                {                    tmp_result.append(s,j,1);                }                else                {                    tmp_maxlen = j - i;                    break;                }            }            if (j == len)            {                tmp_maxlen = j - i ;            }            //cout << "maxlen:" << maxlen << " tmp_maxlen:" << tmp_maxlen << endl;                if (tmp_maxlen > maxlen){                maxlen = tmp_maxlen;            }                //cout << "tmp_result:" << tmp_result << endl;            tmp_result.clear();        }        return maxlen;    }};

方法1.1 （借鉴方法2的思路）：

#include<iostream>#include<algorithm>#include<map>#include<string>using namespace std;int main(){map<char,int> map;string s;cin >> s;int i = 0, maxlen = 0, start = 0;for(i = 0; i< s.length(); i++)// 遍历{if (map.find(s[i]) == map.end()) //子序列中没有重复的{cout << "not find i:" << i << ",start:"<< start <<",maxlen" << maxlen << endl;map[s[i]] = i; // addmaxlen = max(maxlen,i-start+1); cout << "update maxlen:" << maxlen << endl;}else{cout << "has find i:" << i << ",start:"<< start <<",maxlen" << maxlen << endl;map.clear();// TODO:时间复杂度 o(lgn)还是O(n)map[s[i]] = i; // add. important.start = i;}}cout <<"maxlen:" << maxlen << endl;return 0;}

     方法2： java

      思路： if $s[j]$ have a duplicate in the range $[i,j)$ with index j'j​′​​, we don't need to increase $i$ little by little. We can skip all the elements in the range [i, j'][i,j​′​​] and let $i$ to be j' + 1j​′​​+1 directly.

public class Solution {    public int lengthOfLongestSubstring(String s) {        int n = s.length(), ans = 0;        Map<Character, Integer> map = new HashMap<>(); // current index of character        // try to extend the range [i, j]        for (int j = 0, i = 0; j < n; j++) {            if (map.containsKey(s.charAt(j))) {                i = Math.max(map.get(s.charAt(j)), i);            }            ans = Math.max(ans, j - i + 1);            map.put(s.charAt(j), j + 1);        }        return ans;    }}

    方法3：  C++

#include<iostream>#include<vector>#include<algorithm>#include<map>#include<string>using namespace std;int main(){map<char,int> map;vector<int> dict(256,-1);// 初始化-1;字符串下标从0string s;cin >> s;int i = 0, maxlen = 0, start = -1;//初始化-1;字符串下表从0for(i = 0; i< s.length(); i++)// 遍历{if (dict[s[i]] > start)//s[i] 在 先前的子序列出现过{start = dict[s[i]];//新一轮计数位置}dict[s[i]] = i; maxlen = max(maxlen,i-start);}cout <<"maxlen:" << maxlen << endl;return 0;}

      总结： 要分析题目，找出特点或者规律. 写出良好的算法   【how？如何做到？简单的就是动笔画一遍处理流程】.  针对方法2的思路, 当在s[j] 在区间[i,j)中找到一个重复的值了，此时长度是j-i，那么在区间[i,j) 中不可能有更大的不重复的子串了。因此，下次直接从j开始扫描。