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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.
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public class Solution { public int lengthOfLongestSubstring(String s) { if(s == null) return 0; int maxLen = 0; Map<Character, Integer> ind = new HashMap<Character, Integer>(); for(int i = 0; i< s.length(); ++i) { Character c = s.charAt(i); Integer location = ind.get(c); if(location == null) { ind.put(c, i); } else { Integer curLen = ind.size(); if(curLen > maxLen) { maxLen = curLen; } Map<Character, Integer> ind2 = new HashMap<Character, Integer>(); for(int j = location+1;j<=i;++j) { ind2.put(s.charAt(j), j); } ind = ind2; } } Integer curLen = ind.size(); if(curLen > maxLen) { maxLen = curLen; } return maxLen; } }
Improved Slow Version:
public class Solution { public int lengthOfLongestSubstring(String s) { if(s == null) return 0; int maxLen = 0; int start = 0; //improvement: instead of constructing a new map. Modify the original map by removing bad part. Map<Character, Integer> ind = new HashMap<Character, Integer>(); for(int i = 0; i< s.length(); ++i) { Character c = s.charAt(i); Integer location = ind.get(c); if(location == null) { ind.put(c, i); } else { Integer newStart = location+1; Integer curLen = ind.size(); if(curLen > maxLen) { maxLen = curLen; } for(int j = start;j<newStart;++j) { ind.remove(s.charAt(j)); //Improvement: remove bad part. } ind.put(c, i); start = newStart; } } Integer curLen = ind.size(); if(curLen > maxLen) { maxLen = curLen; } return maxLen; } }
More Improvements: No need to remove the bad part.
public class Solution { public int lengthOfLongestSubstring(String s) { if(s == null) return 0; 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){ Character c = s.charAt(i); Integer location = map.get(c); if (location != null){ j = Math.max(j, location + 1); //j points to START. } map.put(c,i); max = Math.max(max,i-j+1); } return max; } }
3. Longest Substring Without Repeating Characters
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原文地址:http://www.cnblogs.com/neweracoding/p/5452212.html