3. Longest Substring Without Repeating Characters

最新推荐文章于 2024-01-06 13:50:14 发布

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最新推荐文章于 2024-01-06 13:50:14 发布

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分类专栏： LeetCode编程

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本文介绍了一种高效算法来解决寻找字符串中最长无重复字符子串的问题，并提供了三种不同的Java实现方案，包括初始版本、优化版本及更优版本。讨论了每种实现的时间复杂度与空间复杂度。

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Description

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given “abcabcbb”, the answer is “abc”, which the length is 3.

Given “bbbbb”, the answer is “b”, with the length of 1.

Given “pwwkew”, 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.

My code :

java:

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length();
        int ans = 0;
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j <= n; j++)
                if (allUnique(s, i, j)) ans = Math.max(ans, j - i);
        return ans;
    }

    public boolean allUnique(String s, int start, int end) {
        Set<Character> set = new HashSet<>();
        for (int i = start; i < end; i++) {
            Character ch = s.charAt(i);
            if (set.contains(ch)) return false;
            set.add(ch);
        }
        return true;
    }
}

But the result is Time Limit Exceeded.

Optimized code:

java:

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length();
        Set<Character> set = new HashSet<>();
        int ans = 0;
        int i = 0, j = 0;
        while (i < n && j < n) {
            if (!set.contains(s.charAt(j))){
                set.add(s.charAt(j++));
                ans = Math.max(ans, j-i);
            }
            else {
                set.remove(s.charAt(i++));
            }
        }
        return ans;
    }
}

Time complexity : O(2n)=O(n).
Space complexity : O(min(m,n)).

Better code:

java:

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