Topics requirements: Given a string, you find out which does not contain a repeated character longest substring length.
Test test case contains:
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: Because the longest substring of characters without repetition is "abc", so its length is 3.
Example 2:
Input: "bbbbb"
Output: 1
Explanation: Because the longest substring is repeated characters without "b", so that its length is 1.
Example 3:
Input: "pwwkew"
Output: 3
Explanation: Because the longest substring is repeated characters without "wke", so its length is 3.
Please note that your answer must be a substring of length, "pwke" is a sub-sequence, not a substring.
I submit verification code is as follows:
The code for the C language version
int lengthOfLongestSubstring(char * s){
int i;
int max = 0;
int left = 0 ; // left border of the window
int table[256];
int tmp=0;
int len = strlen (s);
for(i=0;i<256;i++){
table[i]=-1;
}
for (I = 0 ; I <len; I ++) {// the control window on the right of the boundary
int index = s[i];
IF (left <= Table [index] && Table [index] <= I) {// determines whether the occurrence of repeated characters
= Table left [index] + . 1 ; // Update the left window boundary value
}
I = left-+ tmp . 1 ; // calculate the current size of the sliding window
? Max = tmp> max tmp: max; // updates the maximum length of the substring
table [index] = i; // record the current character in the string index
}
return max;
}
Problem solution ideas:
In the title main use of the concept of sliding window, for the first string, we need to do is to find the maximum length of the substring by a traversal, enabling time complexity is O (N) algorithm to achieve,
在此过程中，将字符串的起始位置设为滑动窗口的左边界，I值作为滑动窗口的右边界，I也就是字符串的下标，在遍历的过程中，我们将当前字符对应的ASCII码
作为了table中的index值，其中的table相当于一个哈希表，每一个元素的value，就是这个字符在字符串中的下标位置，在遍历的时候，每遍历一个字符，我们
都需要计算当前的窗口大小，然后拿当前窗口大小与max值做比较，进行更新，现在问题的关键是，如何控制滑动串口左边界的值，因为滑动窗口的右边界是实时变化的
随着I的增加，可以看到，当发现当前字符在table中对应的值，在窗口之间的时候，就可以判断，发生了字符的重复，需要去更新滑动窗口的左值，那么如何更新呢？
直接将发生碰撞的那个字符在table中的value取出，然后，将value+1赋给left，就是新的左边窗口大小了；在遍历的时候，只要将右边界减去左边界加1就可以计算出当前
窗口大小，然后对最大子串的长度进行更新。
遇到的问题：
1：比较弱智的是，当前在对int型的数组进行初始化的时候，写的是 int table[256]={-1},想要初始化为-1，只能for循环对于C语言而言；
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/longest-substring-without-repeating-characters
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