No.003 Longest Substring Without Repeating Characters
Longest Substring Without Repeating Characters
- Total Accepted: 167158
- Total Submissions: 735821
- Difficulty: Medium
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.
解题思路参考自: http://www.geeksforgeeks.org/length-of-the-longest-substring-without-repeating-characters/
public class Num3 {
/*
* 方法一:暴力搜索,复杂度 O(n^3)
*/
public int lengthOfLongestSubstring(String s) {
if(s == null || s.length() == 0){
return 0 ;
}
String sub ;
for(int subLen = s.length() ; subLen > 0 ; subLen--){
for(int startIndex = 0 ; startIndex <= (s.length()-subLen) ; startIndex++){
//列出所有子串,然后判断子串是否满足有重复
if(startIndex != (s.length()-subLen)){
sub = s.substring(startIndex, startIndex+subLen) ;
}else{
sub = s.substring(startIndex) ;
}
if(!isRepeat(sub)){
return subLen ;
}
}
} return 1 ;
} private boolean isRepeat(String s){
for(int i = 1 ; i < s.length(); i++){
if(s.substring(i).contains(s.substring(i-1, i))){
return true ;
}
}
return false ;
} /*
* 方法二:用hash的方法加上动态规划求解
*/
public int lengthOfLongestSubstring2(String s) {
if(s == null || s.length() == 0){
return 0 ;
}
int cur_len = 1 ; //lenght of current substring
int max_len = 1 ;
int prev_index ; // previous index
int [] visited = new int [256] ;
char [] arr = s.toCharArray() ;
/* Initialize the visited array as -1, -1 is used to
indicate that character has not been visited yet. */
for(int i = 0 ; i < 256 ; i++){
visited[i] = -1 ;
}
/* Mark first character as visited by storing the index
of first character in visited array. */
visited[arr[0]] = 0 ; /* Start from the second character. First character is
already processed (cur_len and max_len are initialized
as 1, and visited[arr[0]] is set */
for(int i = 1 ; i < arr.length ; i++){
prev_index = visited[arr[i]] ; /* If the current character is not present in the
already processed substring or it is not part of
the current NRCS, then do cur_len++ */
if(prev_index == -1 || i - cur_len > prev_index){
cur_len++ ;
}else{
/* Also, when we are changing the NRCS, we
should also check whether length of the
previous NRCS was greater than max_len or
not.*/
if(cur_len > max_len){
max_len = cur_len ;
}
// update the index of current character
cur_len = i - prev_index ;
} visited[arr[i]] = i ;
} // Compare the length of last NRCS with max_len and
// update max_len if needed
if (cur_len > max_len){
max_len = cur_len ;
} return max_len ; } }
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