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).
Example:
Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"
Note:
- If there is no such window in S that covers all characters in T, return the empty string
"". - If there is such window, you are guaranteed that there will always be only one unique minimum window in S.
题意：
给定字符串S 和 T， 求S中可以cover T所有元素的子集的最小长度。
Solution1： Two Pointers(sliding window)
1. scan String T, using a map to record each char's frequency
2. use [leftMost to i] to maintain a sliding window, making sure that each char's frequency in such sliding window == that in T
3. if mapS [S.charAt(start)] > mapT [S.charAt(start)] , it signs we can move sliding window
4. how to find the next sliding window? move leftMost, meanwhile, decrement mapS [S.charAt(start)] until we find each frequency in [start to i] == that in T
code
class Solution {
public String minWindow(String S, String T) {
String result = "";
if (S == null || S.length() == 0) return result;
int[] mapT = new int[256];
int[] mapS = new int[256];
int count = 0;
int leftMost = 0;
for(int i = 0; i < T.length(); i++){
mapT[T.charAt(i)] ++;
}
for(int i = 0; i < S.length(); i++){
char s = S.charAt(i);
mapS[s]++;
if(mapT[s] >= mapS[s]){
count ++;
}
if(count == T.length()){
while(mapS[S.charAt(leftMost)] > mapT[S.charAt(leftMost)]){
if(mapS[S.charAt(leftMost)] > mapT[S.charAt(leftMost)]){
mapS[S.charAt(leftMost)]--;
}
leftMost ++;
}
if(result.equals("") || i - leftMost + 1 < result.length()){
result = S.substring(leftMost, i+1);
}
}
}
return result;
}
}
二刷：
对于出现在S但不出现在T的那些“配角” character的处理，
最好的方式是，边扫S边用map将其频率一并记上。
这样，在判断 while(mapS[S.charAt(leftMost)] > mapT[S.charAt(leftMost)]) 这个逻辑的时候，
这些“配角”character会因为只出现在S但不出现在T
而直接被left++给做掉
class Solution {
public String minWindow(String s, String t) {
String result = "";
if(s == null || s.length() == 0 || s.length() < t.length()) return result;
int [] mapT = new int [128];
for(Character c : t.toCharArray()){
mapT[c]++;
}
int left = 0;
int count = t.length();
int[] mapS = new int[128];
for(int i = 0; i < s.length(); i++){
char c = s.charAt(i);
mapS[c] ++ ;
if(mapT[c] >= mapS[c]){
count --;
}
if(count == 0){
while(mapS[s.charAt(left)] > mapT[s.charAt(left)]){
mapS[s.charAt(left)] --;
left++;
}
if (result.equals("") || i - start + 1 < result.length()) {
result = s.substring(start, i + 1);
}
}
}
return result;
}
}