Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3. Example 2:
Input: "bbbbb" Output: 1 Explanation: The answer is "b", with the length of 1.
Example 3:
Input: "pwwkew" Output: 3 Explanation: 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.
给定一个字符串，请你找出其中不含有重复字符的 最长子串 的长度。
示例 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。
示例 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。
示例 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。
//无重复字符的最长子串 主要是双指针,字符串操作,哈希表
NSString *string = @"ssasa";
NSInteger letfIndicator = 0;
NSInteger longest = 0;
NSMutableSet *strSet = [NSMutableSet set];
//字符串转换为单字符数组
NSMutableArray *charsArray = @[].mutableCopy;
for (NSInteger i = 0; i < string.length; i++) {
if (i < string.length) {
NSString *str = [string substringWithRange:NSMakeRange(i,1)];
[charsArray addObject:str];
}
}
NSLog(@"charsArray = %@",charsArray);
for (int i = 0; i < [charsArray count]; i++) {
NSString *indexStr = charsArray[i];
if ([strSet containsObject:indexStr]) {
longest = MAX(longest,(i - letfIndicator));
while (![charsArray[letfIndicator] isEqualToString:indexStr]) {
[strSet removeObject:charsArray[letfIndicator]];
letfIndicator++;
}
letfIndicator++;
}else{
[strSet addObject:indexStr];
}
}
NSInteger maxLength = MAX(longest,[charsArray count] - letfIndicator);
NSLog(@"maxLength = %ld",maxLength);