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.
要求复杂度为O(n)，只要扫描一遍数组即可，大致做法是设一个start和一个end表示窗口的起始与结束位置，先从先向后匹配end，当找到可容纳T的窗口后再从前而匹配start，去掉无用的元素。具体做法为使用两个哈希表hasCvd与need2Cv，分别表示已经找到的某个字母的匹配与需要找到的匹配。
1 class Solution { 2 public: 3 string minWindow(string S, string T) { 4 int start = 0, end = 0; 5 int min_start = 0, min_end = 0; 6 int min_len = INT_MAX; 7 vector<int> hasCvd(256 ,0); 8 vector<int> need2Cv(256, 0); 9 for (int i = 0; i < T.length(); ++i) { 10 ++need2Cv[T[i]]; 11 } 12 int validCnt = 0; 13 for (end = 0; end < S.length(); ++end) { 14 ++hasCvd[S[end]]; 15 if (need2Cv[S[end]] == 0) continue; 16 if (hasCvd[S[end]] <= need2Cv[S[end]]) ++validCnt; 17 if (validCnt == T.length()) { 18 while(hasCvd[S[start]] > need2Cv[S[start]]) { 19 --hasCvd[S[start]]; 20 ++start; 21 } 22 if(min_len > (end - start)) { 23 min_len = end - start + 1; 24 min_start = start; 25 min_end = end; 26 } 27 } 28 } 29 return (min_len == INT_MAX) ? "" : S.substr(min_start, min_len); 30 } 31 };