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 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"
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