1. 程式人生 > >Given a string, find the length of the longest substring without repeating characters.(給定一個字符串,找到最長的子串的長度,這個子串不存在重復的字符。 )

Given a string, find the length of the longest substring without repeating characters.(給定一個字符串,找到最長的子串的長度,這個子串不存在重復的字符。 )

長度 index val color arraylist pub 翻譯 buffer int

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

翻譯

給定一個字符串,找到最長的子串的長度,這個子串不存在重復的字符。
例如:
輸入”abcabcbb”, 符合條件的子串就是”abc”, 長度為3。

輸入”bbbbb”, 符合條件的子串就是”b”, 長度為1

輸入”pwwkew”, 符合條件的子串就是”wke”, 長度為3。提示:要求必須是子串,例如”pwke” 是一個子序列但不是一個子串。

分析

先把給定的字符串進行處理,把字符串不存在重復的字符都找出來,放進map中,然後遍歷找出最長的返回

代碼:

 1 class Solution {
 2     public int lengthOfLongestSubstring(String s) {
 3         if(null!=s){
 4             
 5             char[] charArray=s.toCharArray();
 6             int len=0;
 7             List<String> list = new
ArrayList<String>(); 8 StringBuffer sb =new StringBuffer(); 9 for(int j=0;j<charArray.length;j++) { 10 11 for(int i=j;i<charArray.length;i++){ 12 if(1==charArray.length) { 13 list.add(String.valueOf(charArray[i])); 14 break; 15 } 16 if(sb.indexOf(String.valueOf(charArray[i]))<0){ 17 18 sb.append(String.valueOf(charArray[i])); 19 }else{ 20 list.add(sb.toString()); 21 22 sb.delete(0,sb.length()); 23 break; 24 } 25 } 26 } 27 28 for(int j=0;j<list.size();j++){ 29 if(list.get(j).length()>len){ 30 len=list.get(j).length(); 31 } 32 33 } 34 return len; 35 } 36 37 return 0; 38 39 } 40 }

Given a string, find the length of the longest substring without repeating characters.(給定一個字符串,找到最長的子串的長度,這個子串不存在重復的字符。 )