• [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

     1 class Solution {
     2     public String minWindow(String S, String T) {
     3         String result = "";
     4         if (S == null || S.length() == 0) return result;
     5         int[] mapT = new int[256];
     6         int[] mapS = new int[256];
     7         int count = 0;
     8         int leftMost = 0;
     9         for(int i = 0; i < T.length(); i++){
    10             mapT[T.charAt(i)] ++;
    11         }
    12         
    13          for(int i = 0; i < S.length(); i++){
    14              char s = S.charAt(i);
    15              mapS[s]++;
    16              if(mapT[s] >= mapS[s]){
    17                  count ++;
    18              }
    19              
    20              if(count == T.length()){ 
    21                  while(mapS[S.charAt(leftMost)] > mapT[S.charAt(leftMost)]){
    22                      if(mapS[S.charAt(leftMost)] > mapT[S.charAt(leftMost)]){
    23                          mapS[S.charAt(leftMost)]--;
    24                      }
    25                      leftMost ++;      
    26                  } 
    27                   if(result.equals("") || i - leftMost + 1 < result.length()){
    28                      result =  S.substring(leftMost, i+1);
    29                  }
    30              }
    31          } 
    32       return result;  
    33     }
    34 }

    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

     1 class Solution {
     2     public String minWindow(String s, String t) {
     3         String result = "";
     4         if(s == null || s.length() == 0 || s.length() < t.length()) return result;
     5         
     6         int [] mapT = new int [128];
     7         for(Character c : t.toCharArray()){  
     8             mapT[c]++;
     9         }
    10         
    11         int left = 0;
    12         int count = t.length();
    13         int[] mapS = new int[128];
    14         for(int i = 0; i < s.length(); i++){ 
    15             char c = s.charAt(i);
    16                 mapS[c] ++ ;
    17                 if(mapT[c] >= mapS[c]){
    18                     count --;      
    19             }
    20             if(count == 0){
    21                 while(mapS[s.charAt(left)] > mapT[s.charAt(left)]){ 
    22                     mapS[s.charAt(left)] --;
    23                     left++;
    24                 }
    25                if (result.equals("") || i - start + 1 < result.length()) { 
    26                     result = s.substring(start, i + 1);
    27                 }
    28             }    
    29         }  
    30      return result;   
    31     }
    32 }
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  • 原文地址:https://www.cnblogs.com/liuliu5151/p/9070180.html
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