标签：tin bst sub answer return hash style 滑动 substring
Given a string, find the length of the longest substring without repeating characters.
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.
1 class Solution { 2 public: 3 int lengthOfLongestSubstring(string s) { 4 if(s.empty()){ 5 return 0; 6 } 7 int max=1; 8 unordered_map<char,int> pos; 9 int start=0; 10 11 for(int i=0;i<s.size();i++){ 12 if(pos.find(s[i])!=pos.end()){ 13 14 int tmp = pos.size(); 15 max=tmp>max?tmp:max; 16 int newstart=pos[s[i]]+1; 17 for(int j=start;j<=pos[s[i]];j++){ 18 pos.erase(s[j]); 19 } 20 start=newstart; 21 } 22 23 pos[s[i]]=i; 24 } 25 26 int tmp = pos.size(); 27 max=tmp>max?tmp:max; 28 return max; 29 } 30 };
grandyang解法一：
滑动窗口问题，用两个指针分别代表窗口的两端，保证窗口里的字母都不相同。
hash table用来存字母上一次出现的index。
1 class Solution { 2 public: 3 int lengthOfLongestSubstring(string s) { 4 int res = 0, left = -1, n = s.size(); 5 unordered_map<int, int> m; 6 for (int i = 0; i < n; ++i) { 7 if (m.count(s[i]) && m[s[i]] > left) { 8 left = m[s[i]]; 9 } 10 m[s[i]] = i; 11 res = max(res, i - left); 12 } 13 return res; 14 } 15 };
grandyang解法二：
1 class Solution { 2 public: 3 int lengthOfLongestSubstring(string s) { 4 vector<int> m(128, -1); 5 int res = 0, left = -1; 6 for (int i = 0; i < s.size(); ++i) { 7 left = max(left, m[s[i]]); 8 m[s[i]] = i; 9 res = max(res, i - left); 10 } 11 return res; 12 } 13 };
3. Longest Substring Without Repeating Characters
标签：tin bst sub answer return hash style 滑动 substring
原文地址：https://www.cnblogs.com/ruisha/p/10204122.html
