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.
思路:问题可以转变为,T中各个字符的数量<= window中的这些字符的数量,所以用map纪录字符与字符数量的关系
class Solution { public: bool ifContain(map<char,int>& source, map<char,int>& target) //check if S contains T { for(map<char,int>::iterator it = target.begin(); it != target.end(); it++) //map的遍历 { if(source[it->first] < it->second) return false; } return true; } string minWindow(string S, string T) { int minLength = INT_MAX; int start = 0, end = 0, minStart = 0, minEnd = 0; map<char,int> source; map<char,int> target; source[S[start]]++; //map的插入[法I]source[key]=value; [法II]source.insert(make_pair(key,value)); for(int i = 0; i< T.length(); i++) { target[T[i]]++; } while(1) { if(ifContain(source, target)){ if(end-start+1 < minLength) { minStart = start; minEnd = end; minLength = end-start+1; if(minLength == T.size()) return S.substr(minStart,minLength); } source[S[start]]--; //寻找更小的窗口 start++; } else //不包含,则扩大窗口 { end++; if(end==S.size()) break; source[S[end]]++; } } if(minLength>S.size()) return ""; else return S.substr(minStart,minLength); } };
时间: 2024-11-03 19:48:34