Timeline
Timeline
2025-11-11
init
Sliding window
Problem:
Sliding window; note the case where the entire string has no repeated characters.
12345678910111213141516171819202122232425262728293031323334 | using std::string;using std::unordered_set;// Given a string s, find the length of the longest substring without repeating characters.class Solution { public: int lengthOfLongestSubstring(string s) { int i = 0, j = 0; int n = s.size(); unordered_set<char> uset; int len = 0; for (j = 0; j < n; j++) { if (uset.count(s[j])) { len = std::max(len, (int)uset.size()); // Shrink the left boundary while (s[i] != s[j]) { uset.erase(s[i]); i++; } if (s[i] == s[j]) { //until the current character is removed uset.erase(s[i]); i++; } } uset.insert(s[j]); } return std::max(len, (int)uset.size()); }}; |
leetcode hot 100 rewrite:
Before entering the loop, shrink the left boundary first until the substring has no repeating characters.
1234567891011121314151617181920212223242526272829303132 | using std::string;using std::unordered_set;class Solution { public: int lengthOfLongestSubstring(string s) { int left = 0, right, n = s.size(); int max_len = 0; unordered_set<char> uset; if (n == 0) return 0; for (right = 0; right < n; right++) { // left index of window // shrink window while (uset.count(s[right])) { uset.erase(s[left]); left++; } uset.insert(s[right]); max_len = std::max(max_len, right - left + 1); } return max_len; }}; |
Template approach:
123456789101112131415161718192021222324252627 | using std::string;using std::unordered_map;class Solution {public: int lengthOfLongestSubstring(string s) { int left = 0, right = 0, n = s.size(); int max_len = 0; unordered_map<char, int> umap; // abcabcbb while (right < n) { umap[s[right]]++; right++; while (umap[s[right - 1]] > 1) { umap[s[left]]--; left++; } max_len = std::max(max_len, right - left); } return max_len; }}; |
