Leetcode 3 Longest Substring Without Repeating Characters. (最长无重复字符子串) (滑动窗口, 双指针)

目录

• 问题描述
• 例子
• 方法一

Leetcode 3

问题描述

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

例子

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.

方法一

  保留一个将字符串中的字符存储为键并将其位置存储为值的hashmap，并保留两个定义最大子字符串的指针。移动右指针以浏览字符串，同时更新hashmap。如果字符已经在hashmap中，则将左指针移到最后找到的相同字符的右边。请注意，两个指针只能向前移动。

** Solution Java **
** 7ms, 79.25% **
** 41.9MB, 5.20% **
class Solution {
    public int lengthOfLongestSubstring(String s) {
        if (s.length() <= 1) return s.length();
        Map<Character, Integer> map = new HashMap<>();
        int cur = 0, max = 0;
        for (int i = 0; i < s.length(); ++i) {
            if (map.containsKey(s.charAt(i))) {
                cur = Math.max(cur, map.get(s.charAt(i)) + 1);
            }
            map.put(s.charAt(i), i);
            max = Math.max(max, i - cur + 1);
        }
        return max;
    }
}

** Solution Python3 **
** 56ms, beats 76.40% **
** 13MB, beats 99.49% **
class Solution:
    def lengthOfLongestSubstring(self, s) :
        cur = maxLength = 0
        usedChar = {}
        for index, char in enumerate(s) :
            if char in usedChar and cur <= usedChar[char]:
                cur = usedChar[char] + 1
            else:
                maxLength = max(maxLength, index - cur + 1)
            usedChar[char] = index
        return maxLength

方法二

** Solution Java **
** 3ms, beats 93.80% **
** 39.1MB, beats 21.47% **
class Solution {
    public int lengthOfLongestSubstring(String s) {
        if (s.length() <= 1) return s.length();
        int[] map = new int[128];
        int ans = 0;
        Arrays.fill(map, -1);
        for (int i = 0, j = 0; j < s.length(); ++j) {
            if (map[s.charAt(j)] >= 0)
                i = Math.max(i, map[s.charAt(j)] + 1);
            ans = Math.max(ans, j - i + 1);
            map[s.charAt(j)] = j;
        }
        return ans;
    }
}