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/
1 public class Num3 { 2 /* 3 * 方法一:暴力搜索,复杂度 O(n^3) 4 */ 5 public int lengthOfLongestSubstring(String s) { 6 if(s == null || s.length() == 0){ 7 return 0 ; 8 } 9 String sub ; 10 for(int subLen = s.length() ; subLen > 0 ; subLen--){ 11 for(int startIndex = 0 ; startIndex <= (s.length()-subLen) ; startIndex++){ 12 //列出所有子串,然后判断子串是否满足有重复 13 if(startIndex != (s.length()-subLen)){ 14 sub = s.substring(startIndex, startIndex+subLen) ; 15 }else{ 16 sub = s.substring(startIndex) ; 17 } 18 if(!isRepeat(sub)){ 19 return subLen ; 20 } 21 } 22 } 23 24 return 1 ; 25 } 26 27 private boolean isRepeat(String s){ 28 for(int i = 1 ; i < s.length(); i++){ 29 if(s.substring(i).contains(s.substring(i-1, i))){ 30 return true ; 31 } 32 } 33 return false ; 34 } 35 36 /* 37 * 方法二:用hash的方法加上动态规划求解 38 */ 39 public int lengthOfLongestSubstring2(String s) { 40 if(s == null || s.length() == 0){ 41 return 0 ; 42 } 43 int cur_len = 1 ; //lenght of current substring 44 int max_len = 1 ; 45 int prev_index ; // previous index 46 int [] visited = new int [256] ; 47 char [] arr = s.toCharArray() ; 48 /* Initialize the visited array as -1, -1 is used to 49 indicate that character has not been visited yet. */ 50 for(int i = 0 ; i < 256 ; i++){ 51 visited[i] = -1 ; 52 } 53 /* Mark first character as visited by storing the index 54 of first character in visited array. */ 55 visited[arr[0]] = 0 ; 56 57 /* Start from the second character. First character is 58 already processed (cur_len and max_len are initialized 59 as 1, and visited[arr[0]] is set */ 60 for(int i = 1 ; i < arr.length ; i++){ 61 prev_index = visited[arr[i]] ; 62 63 /* If the current character is not present in the 64 already processed substring or it is not part of 65 the current NRCS, then do cur_len++ */ 66 if(prev_index == -1 || i - cur_len > prev_index){ 67 cur_len++ ; 68 }else{ 69 /* Also, when we are changing the NRCS, we 70 should also check whether length of the 71 previous NRCS was greater than max_len or 72 not.*/ 73 if(cur_len > max_len){ 74 max_len = cur_len ; 75 } 76 // update the index of current character 77 cur_len = i - prev_index ; 78 } 79 80 visited[arr[i]] = i ; 81 } 82 83 // Compare the length of last NRCS with max_len and 84 // update max_len if needed 85 if (cur_len > max_len){ 86 max_len = cur_len ; 87 } 88 89 return max_len ; 90 91 } 92 93 }
时间: 2024-10-23 00:40:13