LeetCode算法入門- Longest Substring Without Repeating Characters-day4

Longest Substring Without Repeating Characters

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.

方法一：暴力破解法：時間復雜度為：0（n^3）

class Solution {public int lengthOfLongestSubstring(String s) {int n = s.length();int ans = 0;for(int i = 0; i < n; i++){for(int j = i+1; j <= n; j++){if(allUnique(s, i, j))//取最大值ans = Math.max(ans,j-i);}}return ans;}public static boolean allUnique(String s, int start, int end){Set<Character> set = new HashSet<>();for(int i = start; i < end; i++){Character ch = s.charAt(i);//通過set數據結構的contains方法來判斷是否重復if(set.contains(ch))return false;set.add(ch);}return true;} }

方法二：基本思路為：i，j都是從0開始，當s.charAt(j)不在set中的時候，j++，同時set.add(s.charAt(j))；當s.charAt(j)在set當中時，i++，同時移除掉s.charAt(i)，注意i+1加到set中不含s.charAt(j)為止。取出ans與j-i之間的最大值即可。
利用set數據結構

class Solution {public int lengthOfLongestSubstring(String s) {int n = s.length();Set<Character> set =new HashSet<>();int ans = 0;int i = 0;int j = 0;while(i < n && j <n){if(!set.contains(s.charAt(j))){set.add(s.charAt(j++));ans = Math.max(ans, j - i);}else{set.remove(s.charAt(i++));}}return ans;} }

總結

以上是生活随笔為你收集整理的LeetCode算法入门- Longest Substring Without Repeating Characters-day4的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。