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 emtpy string "".
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
采用双指针，end指针逐渐增加，当begin与end之间包含所有t后，后移begin，找到最小substr，并记录，最后输出最小substr。边界条件，begin处的字母属于t且在该区域中仅出现t中的次数。
1 class Solution { 2 public: 3 string minWindow(string s, string t) { 4 int slen=s.size(); 5 int tlen=t.size(); 6 if(tlen==0||slen<tlen) return ""; 7 int needfind[256]={0}; 8 int hasfind[256]={0}; 9 for(int i=0;i<tlen;i++) 10 { 11 needfind[t[i]]++; 12 } 13 int count=0; 14 int begin=0; 15 int end=0; 16 int minwindowsizetmp=0; 17 int minbegin=0; 18 int minend=slen-1; 19 int minwindowsize=INT_MAX; 20 for(count=0;end<slen;end++) 21 { 22 if(needfind[s[end]]==0) 23 continue; 24 hasfind[s[end]]++; 25 if(hasfind[s[end]]<=needfind[s[end]]) 26 { 27 28 count++; 29 } 30 31 if(count==tlen) 32 { 33 while(begin<end) 34 { 35 if(needfind[s[begin]]==0) 36 { 37 begin++; 38 continue; 39 } 40 if(hasfind[s[begin]]>needfind[s[begin]]) 41 { 42 hasfind[s[begin]]--;//若该行放到begin++下边，则会出现下述错误，谨记。 43 begin++; 44 45 continue; 46 } 47 else 48 break; 49 } 50 minwindowsizetmp=end-begin+1; 51 52 if(minwindowsizetmp<minwindowsize) 53 { 54 minbegin=begin; 55 minend=end; 56 minwindowsize=minwindowsizetmp; 57 } 58 } 59 } 60 if(minwindowsize==INT_MAX) 61 return ""; 62 return s.substr(minbegin,minwindowsize); 63 } 64 };
wrong answer:
Input:"ADOBECODEBANC", "ABC"
Output:"ANC"
Expected:"BANC"