LeetCode——Longest Substring Without Repeating Characters

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.

解法一

用HashMap来保存字符与字符出现的位置，用left和i来管理一个滑动窗口，如果当前遍历到的值在窗口内出现过，则更新left指向这个值出现过的位置，然后比较max与i-left的大小决定是否要更新max，再更新当前这个字符的新位置；如果未出现过这个字符则将其加入map。

public int lengthOfLongestSubstring(String s) {
		if(s==null||s.length()==0)
			return 0;
		int left=-1;
		int max=0;
		HashMap<Character,Integer> map=new HashMap<Character, Integer>();
		for(int i=0;i<s.length();i++)
		{
			if(map.containsKey(s.charAt(i))&&map.get(s.charAt(i))>left)
			{
				left=map.get(s.charAt(i));
				map.replace(s.charAt(i), i);
			}
			else
			{
				map.put(s.charAt(i), i);
			}
			max=max>i-left?max:i-left;	
		}
        
		return max;
    }

Runtime: 23 ms, faster than 88.72% of Java online submissions for Longest Substring Without Repeating Characters.
Memory Usage: 39.8 MB, less than 15.92% of Java online submissions for Longest Substring Without Repeating Characters.\

解法二

核心思想与方法一类似，但是用一个长度为256的数组来代替HashMap，是的代码更加简洁

public int lengthOfLongestSubstring(String s) {
		int[] m=new int[256];
		Arrays.fill(m, -1);
		int max=0,left=-1;
		for(int i=0;i<s.length();i++)
		{
			left=Math.max(left, m[s.charAt(i)]);
			m[s.charAt(i)]=i;
			max=Math.max(max, i-left);
				
		}
		return max;
	}

Runtime: 15 ms, faster than 100.00% of Java online submissions for Longest Substring Without Repeating Characters.
Memory Usage: 40.1 MB, less than 13.33% of Java online submissions for Longest Substring Without Repeating Characters.
参考

解法三

用HashSet保存字符，如果当前字符存在于set中了，那么就删除重复的字符和重复字符之前的所有字符

public int lengthOfLongestSubstring(String s) {
		int max=0,left=0;
		HashSet<Character> t=new HashSet<Character>();
		for(int i=0;i<s.length();i++)
		{
			if(!t.contains(s.charAt(i)))
			{
				t.add(s.charAt(i));
				max=Math.max(max, t.size());
			}
			else
			{
				while(t.contains(s.charAt(i)))
					t.remove(s.charAt(left++));
				t.add(s.charAt(i));
			}
		}
		return max;
	}

Runtime: 24 ms, faster than 85.59% of Java online submissions for Longest Substring Without Repeating Characters.
Memory Usage: 39.9 MB, less than 15.36% of Java online submissions for Longest Substring Without Repeating Characters.