[LeetCode] 003. Longest Substring Without Repeating Characters (Medium) (C ++/Java/Python), leetcoderepeating

[LeetCode] 003. Longest Substring Without Repeating Characters (Medium) (C ++/Java/Python), leetcoderepeating

Index: [LeetCode] Leetcode index (C ++/Java/Python/SQL)
Github: https://github.com/illuz/leetcode

003. Longest_Substring_Without_Repeating_Characters (Medium) Link:

Title: https://oj.leetcode.com/problems/Longest-Substring-Without-Repeating-Characters/
Code (github): https://github.com/illuz/leetcode

Question:

From the title, you can understand the meaning of the question. It is to find the longest substring in a string that does not contain repeated characters.

Analysis:

Open an array to record the location where the current character has appeared recently. Calculate it again, update the left boundary, and use it to calculate the maximum value.
Space that requires constant.

Code:

C ++:

class Solution {public:    int lengthOfLongestSubstring(string s) {       int maxlen = 0, left = 0;   int sz = s.length();   int prev[N];   memset(prev, -1, sizeof(prev));   for (int i = 0; i < sz; i++) {   if (prev[s[i]] >= left) {   left = prev[s[i]] + 1;   }   prev[s[i]] = i;   maxlen = max(maxlen, i - left + 1);   }   return maxlen;    }};

Java:

public class Solution {    public int lengthOfLongestSubstring(String s) {        int res = 0, left = 0;        int prev[] = new int[300];// init prev array        for (int i = 0; i < 300; ++i)            prev[i] = -1;        for (int i = 0; i < s.length(); ++i) {            if (prev[s.charAt(i)] >= left)                left = prev[s.charAt(i)] + 1;            prev[s.charAt(i)] = i;            if (res < i - left + 1)                res = i - left + 1;        }        return res;    }}

Python:

class Solution:    # @return an integer    def lengthOfLongestSubstring(self, s):        res = 0        left = 0        d = {}        for i, ch in enumerate(s):            if ch in d and d[ch] >= left:                left = d[ch] + 1            d[ch] = i            res = max(res, i - left + 1)        return res