[Algorithm] 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.

Solution:

In the naive approaches, we repeatedly check a substring to see if it has duplicate character. But it is unnecessary. If a substring $s_{ij}sij from index ii to j - 1j−1 is already checked to have no duplicate characters. We only need to check if s[j]s[j] is already in the substring s_{ij}sij.$

To check if a character is already in the substring, we can scan the substring, which leads to an $O(n^2)O(n2) algorithm. But we can do better.$

By using HashSet as a sliding window, checking if a character in the current can be done in

A sliding window is an abstract concept commonly used in array/string problems. A window is a range of elements in the array/string which usually defined by the start and end indices, i.e.

Back to our problem. We use HashSet to store the characters in current window

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function(s) {
    let begin = 0, max = 0;
    let hash = new Set();
    
    for (let end = 0; end < s.length; end++) {
        if (hash.has(s[end])) {
            while (s[begin] !== s[end]) {
                // delete chars until the dulpicate one
                hash.delete(s[begin++]);
            }
            // delete dulpicate one
            hash.delete(s[begin++]);
        }
        
        hash.add(s[end])
        max = Math.max(max, hash.size)

    }
    return max;
};

[Algorithm] Longest Substring Without Repeating Characters?

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