133 lines
2.4 KiB
Markdown
133 lines
2.4 KiB
Markdown
# Leetcode Longest-Substring-Without-Repeating-Characters
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#### 2022-07-14 09:33
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> ##### Algorithms:
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>
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> #algorithm #Kadane_s_algorithm #sliding_window
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>
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> ##### Data structures:
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>
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> #DS #string
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>
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> ##### Difficulty:
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>
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> #coding_problem #difficulty-medium
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>
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> ##### Additional tags:
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>
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> #leetcode
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>
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> ##### Revisions:
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>
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> N/A
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/longest-substring-without-repeating-characters/)
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---
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### Problem
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Given a string `s`, find the length of the **longest substring** without repeating characters.
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#### Examples
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**Example 1:**
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**Input:** s = "abcabcbb"
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**Output:** 3
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**Explanation:** The answer is "abc", with the length of 3.
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**Example 2:**
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**Input:** s = "bbbbb"
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**Output:** 1
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**Explanation:** The answer is "b", with the length of 1.
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**Example 3:**
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**Input:** s = "pwwkew"
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**Output:** 3
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**Explanation:** The answer is "wke", with the length of 3.
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Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
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#### Constraints
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- `0 <= s.length <= 5 * 104`
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- `s` consists of English letters, digits, symbols and spaces.
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### Thoughts
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> [!summary]
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> This is a #Kadane_s_algorithm
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Initially, I thought of a kadane's algo, but implemented wrong.
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Then I figured out kadane's algorithm.
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similar to this one.
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[[Leetcode Best-Time-To-Buy-And-Sell-Stock]]
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The goal is making cur as small as possible, without duplicating
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And the localMax is max(localMax, i - cur)
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### Solution
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Kadane's algorithm
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```cpp
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class Solution {
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public:
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int lengthOfLongestSubstring(string s) {
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// Kadane's algorithm
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vector<int> hMap(255, -1);
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int cur = -1;
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int localMax = 0;
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for (int i = 0; i < s.length(); i++) {
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if (hMap[s[i]] > cur) {
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// If the element occurs again, reset
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cur = hMap[s[i]];
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}
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hMap[s[i]] = i;
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// The char at cur is ignored
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localMax = max(localMax, i - cur);
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}
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return localMax;
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}
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};
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```
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Initial solution
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```cpp
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class Solution {
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public:
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int lengthOfLongestSubstring(string s) {
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unordered_map<char, int> hMap;
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int cur = 0;
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int localMax = 0;
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for (int i = 0; i < s.size(); i++) {
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if (hMap.find(s[i]) == hMap.end()) {
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cur++;
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localMax = max(cur, localMax);
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hMap[s[i]] = i;
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} else {
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i = hMap[s[i]];
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hMap.clear();
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cur = 0;
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}
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}
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return localMax;
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}
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};
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```
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