leetCode 76. Minimum Window Substring

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,
S = “ADOBECODEBANC”
T = “ABC”
Minimum window is “BANC”.

Note:
If there is no such window in S that covers all characters in T, return the empty string “”.

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

找到s中最短的包含t中所有字符的子串。
思路：
两个标志位start、end，先将end右移，找到最左边的符合条件的子串
然后尝试将start右移，在满足条件的情况下找到最小的串
然后每次将end右移一位，如果不是t中字符，再右移；如果是t中字符，尝试将start右移
用一个数组维护每个字符出现的次数，并用count保存当前最小距离

 public static String minWindow(String s, String t) {         if(s==null)return null;         if(t==null||t.equals(""))return "";         int []A = new int[128];         int [] S = new int[128];         int count=0;         for(int i=0;i<t.length();i++){             A[t.charAt(i)]++;             count++;         }         int count2=0;         for(int i=0;i<s.length();i++){             if(S[s.charAt(i)]<A[s.charAt(i)]){                 S[s.charAt(i)]++;                 count2++;             }             else if(A[s.charAt(i)]>0){                 S[s.charAt(i)]++;             }             if(count2==count)break;         }         if(count2<count)return "";         count2=0;        String res="";        S = new int [128];         int l=0;         int r = 0;         //先找到第一个合适的子串         while(r<s.length()){             char c= s.charAt(r);               if(S[c]<A[c]){                 S[c]++;                 count2++;             }             else if(A[c]!=0)                 S[c]++;             if(count2==count)                 break;             r++;         }         res = s.substring(l,r+1);        System.out.println(res);         int ll=l,rr=r;         //看l能不能右移         while(l<r){             char c= s.charAt(l);              if(A[c]==0){//无关字符，直接移                 l++;                 if(r-l<rr-ll){                     res = s.substring(l,r+1);                     rr=r;                     ll=l;                 }                // System.out.println(s.substring(l,r+1));             }             else if(S[c]>A[c]){//相关字符，计数-1                 l++;                 S[c]--;                 if(r-l<rr-ll){                     res = s.substring(l,r+1);                     rr=r;                     ll=l;                 }                 //System.out.println(s.substring(l,r+1));             }             else break;         }         //每次把r往右移一步，看l能不能右移         r++;         while(r<s.length()){             char c= s.charAt(r);              if(A[c]>0){//找到一个符合条件的字符                S[c]++;                //看l能不能右移                 while(l<r){                     c= s.charAt(l);                      if(A[c]==0){//无关字符，直接移                         l++;                         if(r-l<rr-ll){                             res = s.substring(l,r+1);                             rr=r;                             ll=l;                         }                        // System.out.println(s.substring(l,r+1));                     }                     else if(S[c]>A[c]){//相关字符，计数-1                         l++;                         S[c]--;                         if(r-l<rr-ll){                             res = s.substring(l,r+1);                             rr=r;                             ll=l;                         }                         //System.out.println(s.substring(l,r+1));                     }                     else break;                 }            }            r++;         }             return res;        }