LeetCode3. Longest Substring Without Repeating Characters

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:

Input: s = "pwwkew"
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.

思路

• 定义一个哈希表存放字符及其出现的位置；
• 定义 i, j 分别表示不重复子串的开始位置和结束位置；
• j 向后遍历，若遇到与[i, j]区间内字符相同的元素，更新 i 的值，此时 [i, j] 区间内不存在重复字符，计算 res 的最大值。

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int res = 0;
        Map<Character, Integer> chars = new HashMap<>();
        for (int i = 0, j = 0; j < s.length(); ++j) {
            char c = s.charAt(j);
            if (chars.containsKey(c)) {
                // chars.get(c)+1 可能比 i 还小，通过 max 函数来锁住左边界
                // e.g. 在"tmmzuxt"这个字符串中，遍历到最后一步时，最后一个字符't'和第一个字符't'是相等的。如果没有 max 函数，i 就会回到第一个't'的索引0处的下一个位置
                i = Math.max(i, chars.get(c) + 1);
            }
            chars.put(c, j);
            res = Math.max(res, j - i + 1);
        }
        return res;
    }
}