花花酱 LeetCode 3. Longest Substring Without Repeating Characters

zxi — Wed, 12 Sep 2018 04:43:30 +0000

Problem

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.

Solution: HashTable + Sliding Window

Using a hashtable to remember the last index of every char. And keep tracking the starting point of a valid substring.

start = max(start, last[s[i]] + 1)

ans = max(ans, i – start + 1)

Time complexity: O(n)

Space complexity: O(128)

C++

// Author: Huahua
class Solution {
public:
  int lengthOfLongestSubstring(string s) {
    const int n = s.length();
    int ans = 0;
    vector idx(128, -1);
    for (int i = 0, j = 0; j < n; ++j) {
      i = max(i,idx[s[j]] + 1);
      ans = max(ans, j - i + 1);      
      idx[s[j]] = j;
    }
    return ans;
  }
};

Java

// Author: Huahua
class Solution {
  public int lengthOfLongestSubstring(String s) {
    int[] last = new int[128];
    Arrays.fill(last, -1);    
    int start = 0;
    int ans = 0;
    for (int i = 0; i < s.length(); ++i) {      
      if (last[s.charAt(i)] != -1)
        start = Math.max(start, last[s.charAt(i)] + 1);
      last[s.charAt(i)] = i;
      ans = Math.max(ans, i - start + 1);      
    }
    return ans;
  }
}

Python3

# Author: Huahua
class Solution:
  def lengthOfLongestSubstring(self, s):
    last = [-1] * 128
    ans = 0
    start = 0
    for i, ch in enumerate(s):
      if last[ord(ch)] != -1:
        start = max(start, last[ord(ch)] + 1)
      ans = max(ans, i - start + 1)
      last[ord(ch)] = i
    return ans

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花花酱 LeetCode 3. Longest Substring Without Repeating Characters

Problem

Solution: HashTable + Sliding Window

C++

Java

Python3