[LeetCode] 3.Longest Substring Without Repeating Characters 最长无重复子串

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.

滑动窗口法，右指针从左往右扫描字符串，遇到没出现过的字符就直接记录，当扫到的字符出现过，左指针一直向右走，直到遇到重复字符时，左指针在右移1位。比较当前不重复字符串长度与max值，更新max值。最后返回max值。

字符的记录可用hashmap或int[256]， 记录和查寻字符的方式有两种：

第一种记录每个字符上次出现的位置，当遇到新字符时，直接记录字符位置。当遇到的字符存在时，如果记录的位置小于左指针的位置，说明被跳过了，重新记录当前位置。当遇到重复的字符存在并且在左指针右边(在窗口内)时，指针直接跳到记录的这个字符上次出现的位置加1，就是跳过这个字符，然后记录这个新的位置。

第二种记录字符是否出现，如果扫到重复的字符，左指针向右移动，到重复字符的位置加1，这之间扫过的字符都删掉，因为不在窗口里了。

Java:

  public int lengthOfLongestSubstring(String s) {
        if (s.length()==0) return 0;
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        int max=0;
        for (int i=0, j=0; i<s.length(); ++i){
            if (map.containsKey(s.charAt(i))){
                j = Math.max(j,map.get(s.charAt(i))+1);
            }
            map.put(s.charAt(i),i);
            max = Math.max(max,i-j+1);
        }
        return max;
    }　　

Java: HashMap

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        HashMap<Character, Integer> map = new HashMap<Character,Integer>();
        int maxLen = 0;
        int start = 0;
        for (int i = 0; i< s.length(); i++) {
            if (!map.containsKey(s.charAt(i)) || map.get(s.charAt(i)) < start) {
                map.put(s.charAt(i),i);
            }
            else {
                maxLen = Math.max(maxLen,i - start);
                start = map.get(s.charAt(i)) + 1;
                map.put(s.charAt(i),i);
            }
        }

        maxLen = Math.max(maxLen, s.length() - start);

        return maxLen;
    }
}

Java 2: HashSet

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int longest = 0;
        int i = 0;
        Set<Character> set = new HashSet<>();

        for (int j = 0; j < s.length(); j++) {
            while (set.contains(s.charAt(j))) {
                set.remove(s.charAt(i));
                i++;
            }

            set.add(s.charAt(j));
            longest = Math.max(longest, j - i + 1);
        }
        return longest;
    }
}

Java 3: boolean[256]

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        boolean[] visited = new boolean[256];
        int start = 0, last = 0, max = 0, length = s.length();
        while(last < length) {
            if(!visited[s.charAt(last)]) {
                visited[s.charAt(last)] = true;
                max = Math.max(max, last++ - start + 1);
            } else {
                while(visited[s.charAt(last)])
                    visited[s.charAt(start++)] = false;
                visited[s.charAt(last++)] = true;
            }
        }
        return max;
    }
}

Java 4: int[256]

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

Python:

class Solution2(object):
    def lengthOfLongestSubstring(self, s):
        a={}
        count=0
        first=-1
        for i in range(len(s)):
            if s[i] in a and a[s[i]]>first:
                first=a[s[i]]
            a[s[i]]=i
            count=max(count,(i-first))
        return count

Python:

class Solution(object):
    def lengthOfLongestSubstring(self, s):
        a=[-1]*256
        count=0
        first=-1
        for i in range(len(s)):
            if a[ord(s[i])]>first:
                first=a[ord(s[i])]
            a[ord(s[i])]=i
            count=max(count,(i-first))
        return count

Python:

class Solution:
    def lengthOfLongestSubstring(self, s):
        longest, start, visited = 0, 0, [False for _ in xrange(256)]
        for i, char in enumerate(s):
            if visited[ord(char)]:
                while char != s[start]:
                    visited[ord(s[start])] = False
                    start += 1
                start += 1
            else:
                visited[ord(char)] = True
            longest = max(longest, i - start + 1)
        return longest

Python: wo

class Solution():
    def longestSubstring(self, s):
        longest, lookup = 0, {}
        for k, v in enumerate(s):
            if v not in lookup:
                lookup[v] = k
                longest = max(longest, len(lookup))
            else:
                for j in range(lookup[v], k):
                    lookup.pop(s[j])

        return longest　　

Python: wo

class Solution(object):
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        start, mx, m = 0, 0, {}
        for i in xrange(len(s)):
            if s[i] in m and m[s[i]] >= start:
                start = m[s[i]] + 1
            m[s[i]] = i
            mx = max(mx, i - start + 1)

        return mx　

C++:

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int m[256] = {0}, res = 0, left = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (m[s[i]] == 0 || m[s[i]] < left) {
                res = max(res, i - left + 1);
            } else {
                left = m[s[i]];
            }
            m[s[i]] = i + 1;
        }
        return res;
    }
};　

C++:

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        vector<int> m(256, -1);
        int res = 0, left = -1;
        for (int i = 0; i < s.size(); ++i) {
            left = max(left, m[s[i]]);
            m[s[i]] = i;
            res = max(res, i - left);
        }
        return res;
    }
};

　　　　

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