leetcode - 3.Longest Substring Without Repeating Characters

最长无重复子串算法

最新推荐文章于 2025-05-18 01:04:12 发布

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最新推荐文章于 2025-05-18 01:04:12 发布

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分类专栏： leetcode题集文章标签： leetcode

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本文介绍了一种求解字符串中最长无重复字符子串的算法，并提供了两种不同的实现方式。通过动态规划思想来高效地解决该问题，适用于多种编程挑战。

Longest Substring Without Repeating Characters

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.

thinking:

使用动态规划的思想，动态计算最大长度

bacdfeaads
|   |     
 ==> 向最大字符串内添加一个字符
bacdfeaads
|    |   
 ==> 下一个字符是重复字符
bacdfeaads
  |  |   
 ==> 将head后移
bacdfeaads
  |   |       
 ==> 添加
...

Solution1:

  public int lengthOfLongestSubstring(String s) {
        int head = 0;
        int tail = 0;
        int longest = 0;
        int length = s.length();
        String now;
        String letter;

        for (; tail < length; tail++) {
            now = s.substring(head, tail);
            letter = s.substring(tail, tail + 1);

            if (now.indexOf(letter) < 0) {
                if (longest < now.length() + 1) {
                    longest = now.length() + 1;
                }
            } else {
                head = s.indexOf(letter, head) + 1;
            }
        }
        return longest;
    }

Soluction2:

答案

  public int lengthOfLongestSubstring1(String s) {
        int[] letters = new int[128];
        int longest = 0;
        int index = 0;

        for(int i = 0; i < s.length(); i++){
            index = Math.max(letters[s.charAt(i)], index);

            longest = Math.max(i - index + 1, longest);
            //+1有深意
            //  1. s可能是"" => longest不能为1
            //  2. s的长度可能是1 => i - index + 1
            //综上两个原因 => i + 1
            letters[s.charAt(i)] = i + 1;
        }

        return longest;
    }