String hash table to solve the problem

Title 5: no repeat of the longest character substring

Given a string, you find out which does not contain a repeated character  longest substring  length.

Example 1:

输入: "abcabcbb"
输出: 3 
解释: 因为无重复字符的最长子串是 "abc"，所以其长度为 3。

Example 2:

输入: "bbbbb"
输出: 1
解释: 因为无重复字符的最长子串是 "b"，所以其长度为 1。

Example 3:

输入: "pwwkew"
输出: 3
解释: 因为无重复字符的最长子串是 "wke"，所以其长度为 3。
     请注意，你的答案必须是 子串 的长度，"pwke" 是一个子序列，不是子串。

 方法一：

@ Idea: all elements of the control zone is not repeated, using a hash table to determine. When emptying each iteration
 // hash table 
class Solution {
 public :
     int lengthOfLongestSubstring ( String S) { 
        
        int len = s.size ();
         int HashMap [ 256 ] = { 0 };
         int In Flag = 0 ;
         int Current = 0 ;
         int Result = Current;
         int I = 0 ;
         for (I = 0 ; I <len; I ++ ) 
        { 
            IF(HashMap [S [I]] =! 0 ) 
            { 
                // this sub-string duplication 
                Current = I- In Flag;
                 IF (Result < Current) 
                    Result = Current;
                 // clears the hash table 
                for (; In Flag <I; In Flag ++ ) 
                { 
                    IF (S [In Flag] =! S [I]) 
                        HashMap [S [In Flag]] = 0 ;
                     the else 
                        BREAK ; 
                } 
                HashMap [S [I]] = . 1 ; 
                In Flag++ ; 
            } 
            the else { 
                HashMap [S [I]] = . 1 ; // set to 0 
            } 
        } 
        // process without overlap. 
        I- = Current In Flag;
         IF (Result < Current) 
            Result = Current; 

        return Result; 
    } 
};

Method Two:

// Method two: sliding window is advanced my approach. Using hash tables to store the position of the last character string appears 
int lengthOfLongestSubstring ( String S) { 
        Vector < int > hash ( 256 , - . 1 );
         int left = - . 1 ;
         int ANS = 0 ;
         for ( int I = 0 ; I <s.size (); ++ I) {
             IF (the hash [S [I]]> left) { 
                left = the hash [S [I]]; 
            } 
            ANS = max (ANS, I- left); 
            the hash [S [i]]= i;
        }
        return ans;
    }

387 title 

The first string a unique character

Given a string, find its first non-repeating characters, and returns its index. If not, it returns -1.

Case:

s = "leetcode"

返回 0.

s = "loveleetcode",
返回 2.

题解：

 题解，使用hash表对字符出现的个数进行统计，之后再次遍历字符串查找hash表即可。注意出现次数问题的都可以用这个思路来求解！

// @lc code=start
class Solution {
public:
    int firstUniqChar(string s) {
        int len = s.size();
        int result = 0;
        int hashmap[256] = {0};
        for(int i=0;i<len;i++)
        {
            hashmap[s[i]]++;
        }
        for(int i=0;i<len;i++)
        {
            if(HashMap [S [I]] == . 1 )
                 return I; 
        } 
        return - . 1 ; 

    } 
}; // do not traverse the hash table, then traverse a string like! ! !