Minimum Window Substring &&& Longest Substring Without Repeating Characters 快慢指针，都不会退，用hashmap或者其他结构保证

 1 

public class Solution {

 2     public static int lengthOfLongestSubstring(String s) {

 3  

 4     char[] arr = s.toCharArray();

 5     int pre = 0;

 6  

 7     HashMap<Character, Integer> map = new HashMap<Character, Integer>();

 8  

 9     for (int i = 0; i < arr.length; i++) {

10         if (!map.containsKey(arr[i])) {

11             map.put(arr[i], i);

12         } else {

13             pre = pre > map.size() ? pre : map.size();

14             i = map.get(arr[i]);

15             map.clear();

16         }

17     }

18  

19     return Math.max(pre, map.size());

20 }

21 }

 

 1 public class Solution {

 2     public String minWindow(String S, String T) {

 3         char s[]=S.toCharArray();

 4         if(S=="")return "";

 5         int beg=0;

 6         int end=0;

 7         int d[]=new  int[128];

 8         int size=0;

 9         for(int i=0;i<T.length();i++)

10         {

11             if(d[t[i]]==0) size++;

12             d[t[i]]++;

13         }

14         int d2[]=new int[128];

15         int s1=0;

16         boolean flag=false;

17         

18         while(end<s.length)

19         {

20             if(d[s[end]]==0) {end++;continue;}

21              d2[s[end]]++; 

22              if(d2[s[end]]==d[s[end]]) s1++;

23              end++;

24              if(s1==size)

25              {

26                  while(d2[s[beg]]>d[s[beg]]||d[s[beg]]==0) {d2[s[beg]]--;beg++;}

27                  flag=true;

28                  break;

29              }

30            

31             

32         }

33        if(!flag) return "";

34         int aend=end-1;

35         int abeg=beg;

36         int amin=end-beg;

37         

38     

39          while(end<s.length)

40          {

41              if(d[s[end]]==0){end++;continue;}

42              d2[s[end]]++;

43             

44             while((d2[s[beg]]>d[s[beg]])||d[s[beg]]==0) { 

45                 d2[s[beg]]--;beg++;

46                 if(end-beg+1<amin)

47                 {amin=end-beg+1;

48                 aend=end;

49                 abeg=beg;

50             

51                 }

52             

53                 

54                 

55                 

56             }

57             end++;

58              

59          }

60         

61             

62         

63         // 

64         return S.substring(abeg,aend+1);

65         

66         

67         

68         

69         

70         

71       

72         

73     }

74 }

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