LeetCode 3 Longest Substring Without Repeating Characters-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_36293509/article/details/114297288

本文介绍了一种使用滑动窗口方法寻找给定字符串中最长无重复字符子串长度的有效算法。通过实例演示如何更新窗口并跟踪最大长度，适用于多种编程挑战。

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

Example 1:

Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

Example 4:

Input: s = ""
Output: 0

Constraints:

0 <= s.length <= 5 * 104
s consists of English letters, digits, symbols and spaces.

/**
 * 滑动窗口
 * 添加了什么元素就弹出左边元素及其之前的元素
 * a b c a b c b b
 * [a b c] 3
 * a [b c a] 3 添加了a,弹出左边的a
 * b [c a b] 3 添加了b,弹出左边的b
 * c [a b c] 3 添加了c,弹出左边的c
 * a b [c b] 2 添加了b,弹出左边b和b之前所有元素
 * c b [b] 1 添加了b,弹出左边b和b之前所有元素
 */
public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if (s.length() == 0) {
            return 0;
        }
        HashMap<Character, Integer> map = new HashMap<>();
        int max = 0;
        int left = 0;

        for (int i = 0; i < s.length(); i++) {
            if (map.containsKey(s.charAt(i))) {
                left = Math.max(left, map.get(s.charAt(i)) + 1);
            }
            map.put(s.charAt(i), i);
            max = Math.max(max, i - left + i);
        }
        return max;
    }
}