【LeetCode 3. Longest Substring Without Repeating Characters】

最新推荐文章于 2021-12-03 15:58:56 发布

millerkevin

最新推荐文章于 2021-12-03 15:58:56 发布

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分类专栏：【LeetCode】

本文链接：https://blog.youkuaiyun.com/MillerKevin/article/details/105739286

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本文探讨了如何在给定字符串中找到最长的无重复字符子串，提供了两种Kotlin语言实现方法，一种使用哈希映射跟踪字符位置，另一种采用滑动窗口策略。通过实例演示了算法的有效性和效率。

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Description：

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.

翻译：

给定一个字符串，找出其中不含有重复字符的最长子串的长度。

Kotlin代码实现①


    public int lengthOfLongestSubstring(String s) {
        Map<Character, Integer> map = new HashMap<Character, Integer>();
        char[] charArr = s.toCharArray();
        int max = 0;
        int count = 0;
        for (int i = 0; i < charArr.length; i++) {
            Integer last = map.get(charArr[i]);
            if (last == null) {
                count++;
                if (count > max) {
                    max = count;
                }
            } else {
                int start=i-count;
                for(int j=start;j<last;j++){
                    map.remove(charArr[j]);
                    count--;
                }
            }
            map.put(charArr[i], i);
        }
        return max;
    }

Kotlin代码实现②

fun lengthOfLongestSubstring(s: String): Int {
    val n = s.length
    val set = HashSet<Char>()
    var max = 0
    var i = 0
    var j = 0
    while (i < n && j < n) {
        if (!set.contains(s[j])) {
            set.add(s[j++])
            max = Math.max(max, j - i)
        } else {
            set.remove(s[i++])
        }
    }
    return max
}