[LeetCode] 3. Longest Substring without Repeating Characters（最长的无重复字符的子串）

• Difficulty: Medium

• Related Topics: Hash Table, Two Pointers, String, Sliding Window.

• Link: https://leetcode.com/problems/longest-substring-without-repeating-characters/

Description

Given a string s, find the length of the longest substring without repeating characters.
给定一字符串 s，返回其最长的无重复子串的长度。

Examples

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 * 1e4
• s consists of English letters, digits, symbols and spaces.

Solution

这种子串类的题目，用滑动窗口可以比较直观地解决。就是维护一个窗口，保证窗口内的元素不重复。具体怎么维护，参见以下代码：

import kotlin.math.max

class Solution {
    fun lengthOfLongestSubstring(s: String): Int {
        var result = 0
        // 滑动窗口的左右边界
        var left = 0
        var right = 0
        // 子串内字符和其下标的映射
        val charIndices = hashMapOf<Char, Int>()

        while (right < s.length) {
            // right 向右扩展，直到遇见重复字符
            while (right < s.length && !charIndices.containsKey(s[right])) {
                charIndices[s[right]] = right
                right++
            }
            // 更新结果
            result = max(result, right - left)
            if (right == s.length) {
                break
            }
            // left 向右扩展，舍弃在重复字符左侧的所有字符
            val dupIndex = charIndices.getValue(s[right])
            while (left <= dupIndex) {
                charIndices.remove(s[left])
                left++
            }
        }

        return result
    }
}