[leetcode]76. Minimum Window Substring最小字符串窗口

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;
     }
 }