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

本文链接：https://blog.youkuaiyun.com/qq_39008204/article/details/79939204

本文介绍了一种解决最长无重复字符子串问题的有效算法。通过遍历字符串并使用数组记录字符位置，该算法能够找到最长的不包含重复字符的子串长度。支持多种编程语言实现，包括C/C++、Java及Python。

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

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

解题思路：题中所描述的字符包括所有128个ASCII字符，定义一个数组alph记录每个字符出现的位置，并且用一个变量mark记录当前子串起点的位置，变量maxlen记录当前最长的子串的长度，字符的位置随着循环不断往后进行更新，如果出现字符的上一个位置在当前子串起点之后，更新起点的位置为当前字符上一个位置的后一位作为起点，同时判断是否需要更新maxlen。

C/C++

int lengthOfLongestSubstring(char* s) {
    int i = -1;
    int alph[128] = {0};
    for(int i=0;i<128;i++)alph[i]=-1;
    int mark = -1;
    int maxlen = 0;
    while(s[i+1])
    {
        i++;
        int tmp = s[i];
        if(alph[tmp]>mark)
        {
            if(maxlen<i-mark-1)maxlen=i-mark-1;
            mark = alph[tmp];
        }
        alph[tmp] = i;
    }
    if(maxlen<i-mark)maxlen=i-mark;
    return maxlen;
}

Java

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int[] alph = new int[128];
        for(int i=0;i<128;i++)alph[i]=-1;
        int i = -1;
        int mark = -1;
        int maxlen = 0;
        while(i+1<s.length()) {
            i++;
            int tmp = s.charAt(i);
            if(alph[tmp]>mark) {
                if(maxlen<i-mark-1)maxlen=i-mark-1;
                mark = alph[tmp];
            }
            alph[tmp] = i;
        }
        if(maxlen<i-mark)maxlen=i-mark;
        return maxlen;
    }
}

Python

class Solution:
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        alph = []
        for i in range(128):alph.append(-1)
        i = -1
        mark = -1
        maxlen = 0
        while i+1<len(s):
            i += 1
            tmp = ord(s[i])#字符转整型要用函数，不可直接赋值
            if alph[tmp]>mark:
                if maxlen<i-mark-1:maxlen=i-mark-1
                mark = alph[tmp]
            alph[tmp] = i;
        if maxlen<i-mark:maxlen=i-mark
        return maxlen