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.
题意是在s中找到包含t中所有元素的最小子串，方法是维护一个hashmap，用两个指针在s上滑动，先滑动右窗口直到左右之间的子串满足条件，再滑动左窗达到最小子串
public class Solution { public String minWindow(String s, String t) { if(s.length()==0)return ""; HashMap<Character,Integer>map = new HashMap<Character,Integer>(); for(int i=0;i<t.length();i++){ if(!map.containsKey(t.charAt(i))){ map.put(t.charAt(i),1); } else map.put(t.charAt(i),map.get(t.charAt(i))+1); } int l = 0; int count = 0; int minlen = s.length()+1; int minstart = 0; for(int r=0;r<s.length();r++){ if(map.containsKey(s.charAt(r))){ map.put(s.charAt(r),map.get(s.charAt(r))-1); if(map.get(s.charAt(r))>=0)count++; while(count==t.length()){ if(r-l+1<minlen){ minlen = r-l+1; minstart = l; } if(map.containsKey(s.charAt(l))){ map.put(s.charAt(l),map.get(s.charAt(l))+1); if(map.get(s.charAt(l))>0)count--; } l++; } } } if(minlen>s.length())return ""; return s.substring(minstart,minstart+minlen); } }
