标签：must bst substr note 方法 == pre map etc
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.
解法一：Brute Force
把每两个字母之间是否有重复都判断一遍，如果没有重复就更新x值。
用一个独立的方法来判断是否有重复字母，用上了HashSet的contains来判断重复性。
但是这个解法要O(n3),时间太长了，所以不是一个好解法。
class Solution { public int lengthOfLongestSubstring(String s) { int x=0; int l = s.length(); for(int i =0; i<l; i++) { for(int j=i+1; j<=l; j++) { if(allUnique(s,i,j)) x = Math.max(x,j-i); } } return x; } public 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); if(set.contains(ch)) return false; set.add(ch); } return true; } }
☆ 解法二：和我最初的想法很相似，很好地将我的那个想法实现了出来
用两个指针i和j，j专门指向重复字母的后一个字母
O（n）
public int lengthOfLongestSubstring(String s) { if (s.length()==0) return 0; HashMap<Character, Integer> map = new HashMap<Character, Integer>(); int max=0; for (int i=0, j=0; i<s.length(); ++i){ if (map.containsKey(s.charAt(i))){ //注意charAt的c是小写！！！ j = Math.max(j,map.get(s.charAt(i))+1); } map.put(s.charAt(i),i); max = Math.max(max,i-j+1); } return max; }
leetcode 3. Longest Substring Without Repeating Characters
标签：must bst substr note 方法 == pre map etc
原文地址：https://www.cnblogs.com/jamieliu/p/10296779.html
