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.
給定一個字符串，請你找出其中不含有重複字符的 最長子串 的長度。
示例 1:
輸入: "abcabcbb" 輸出: 3 解釋: 因爲無重複字符的最長子串是 "abc"，所以其長度爲 3。
示例 2:
輸入: "bbbbb" 輸出: 1 解釋: 因爲無重複字符的最長子串是 "b"，所以其長度爲 1。
示例 3:
輸入: "pwwkew"
輸出: 3
解釋: 因爲無重複字符的最長子串是 "wke"，所以其長度爲 3。
請注意，你的答案必須是 子串 的長度，"pwke" 是一個子序列，不是子串。
//無重複字符的最長子串 主要是雙指針,字符串操作,哈希表
NSString *string = @"ssasa";
NSInteger letfIndicator = 0;
NSInteger longest = 0;
NSMutableSet *strSet = [NSMutableSet set];
//字符串轉換爲單字符數組
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);