【LeetCode-3】Longest Substring Without Repeating Characters_find the lenth of the longest substring without re

时隔4年，继续记录一下自己的成长历程吧。
题目：medium
Given a string s, find the length of the longest substring without repeating characters.

Example 1:
Input: s = “abcabcbb”
Output: 3
Explanation: The answer is “abc”, with the length of 3.

Example 2:
Input: s = “bbbbb”
Output: 1
Explanation: The answer is “b”, with the length of 1.

Example 3:
Input: s = “pwwkew”
Output: 3
Explanation: The answer is “wke”, with the length of 3.
Notice that the answer must be a substring, “pwke” is a subsequence and not a substring.

Example 4:
Input: s = “”
Output: 0

Constraints:
0 <= s.length <= 5 * 104
s consists of English letters, digits, symbols and spaces.

这道题的目标是找到给定字符串s的最长无重复字符子串的长度。这里只要求输出最长长度，不要求输出最长子串。
15年(一刷): O(N)

    public int lengthOfLongestSubstring(String s) {
		int res = 0, left = 0;  
        int prev[] = new int[300];  
  
        // init prev array  
        for (int i = 0; i < 300; ++i)  
            prev[i] = -1;  
  
        for (int i = 0; i < s.length(); ++i) {  
            if (prev[s.charAt(i)] >= left)  
                left = prev[s.charAt(i)] + 1;  
            prev[s.charAt(i)] = i;  
            if (res < i - left + 1)  
                res = i - left + 1;  
        }  
        return res;
	}
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19年(二刷): O(N)

   public int lengthOfLongestSubstring(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }
            
        Set<Character> set = new HashSet<Character>();
        char[] strArray = s.toCharArray();
        int result = 1, begin = 0;
        for(int i = 0, len = strArray.length; i < len; i++) {
            if(!set.contains(strArray[i])) {
                set.add(strArray[i]);
                result = Math.max(result, i - begin + 1);
            } else {
                while(strArray[begin] != strArray[i]) {
                    set.remove(strArray[begin]);
                    begin += 1;
                }
                begin += 1;
            }
        }
        
        return result;
    }
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20年(三刷): O(N)

    public int lengthOfLongestSubstring(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }

        int left = 0, right = 0, len = s.length(), result = 1;
        Set<Character> set = new HashSet<>();
        while (left < len && right < len) {
            if (!set.contains(s.charAt(right))) {
                set.add(s.charAt(right));
                right += 1;
                result = Math.max(result, right - left);
            } else {
                set.remove(s.charAt(left));
                left += 1;
            }
        }

        return result;
    }
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