LeetCode 76: Minimum Window Substring

 1 public class Solution {
 2     public String minWindow(String s, String t) {
 3         if (s.length() == 0 || s.length() < t.length() || t.length() == 0) {
 4             return "";
 5         }
 6         Map<Character, Integer> charMap = new HashMap<>();
 7         for (char c : t.toCharArray()) {
 8             charMap.put(c, charMap.getOrDefault(c, 0) + 1);
 9         }
10         int totalCount = 0;
11         int length = Integer.MAX_VALUE;
12         int startPoint = 0;
13         
14         
15         for (int left = 0, right = 0; right < s.length(); right++) {
16             if (charMap.containsKey(s.charAt(right))) {
17                 charMap.put(s.charAt(right), charMap.get(s.charAt(right)) - 1);
18                 if (charMap.get(s.charAt(right)) == 0) {
19                     totalCount++;
20                 }
21             }   
22             
23             while (totalCount == charMap.size()) {
24                 if (length > right - left + 1) {
25                     length = right - left + 1;
26                     startPoint = left;
27                 }
28                 
29                 if (charMap.containsKey(s.charAt(left))) {
30                     charMap.put(s.charAt(left), charMap.get(s.charAt(left)) + 1);
31                     if (charMap.get(s.charAt(left)) > 0) {
32                         totalCount--;
33                     }
34                 }
35                 left++;
36             }
37         }
38         return length == Integer.MAX_VALUE ? "" : s.substring(startPoint, startPoint + length);
39     }
40 }

FOLLOW UPS:

What if the input is coming streaming and you cannot even store the window? But you can store the checked String. I purposed this algorithm that store the index of the second String since the chars that not in the second String are useless. You can skipp them.

 1 public class Solution {
 2   private int length;
 3 
 4   public String minWindow(String s, String t) {
 5     if (s.length() == 0 || s.length() < t.length() || t.length() == 0) {
 6       return "";
 7     }
 8     Map<Character, LinkedList<Integer>> charMap = new HashMap<>();
 9     Map<Character, Integer> countMap = new HashMap<>();
10 
11     for (char c : t.toCharArray()) {
12       countMap.put(c, countMap.getOrDefault(c, 0) + 1);
13     }
14 
15     String result = "";
16     length = Integer.MAX_VALUE;
17     int total = 0;
18 
19     for (int i = 0; i < s.length(); i++) {
20       char current = s.charAt(i);
21       if (countMap.containsKey(current)) {
22         if (!charMap.containsKey(current)) {
23           charMap.put(current, new LinkedList<>());
24         }
25         LinkedList q = charMap.get(current);
26         q.offer(i);
27 
28         if (q.size() > countMap.get(current)) {
29           q.poll();
30         }
31 
32         charMap.put(current, q);
33 
34         result = checkMax(charMap, countMap, result, s);
35       }
36     }
37     return result;
38   }
39 
40   private String checkMax(Map<Character, LinkedList<Integer>> charMap, Map<Character, Integer> countMap, String result,
41       String s) {
42     int min = Integer.MAX_VALUE;
43     int max = -1;
44     for (char c : countMap.keySet()) {
45       if (!charMap.containsKey(c) || charMap.get(c).size() != countMap.get(c)) {
46         return "";
47       }
48 
49       min = Math.min(min, charMap.get(c).getFirst());
50       max = Math.max(max, charMap.get(c).getLast());
51     }
52 
53     if (max - min + 1 < length) {
54       length = max - min + 1;
55       return s.substring(min, max + 1);
56     }
57     return length == Integer.MAX_VALUE ? "" : result;
58   }
59 }

posted on 2017-07-23 08:21 keepshuatishuati 阅读(77) 评论(0) 编辑收藏举报