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LeetCode: Minimum Window Substring
1 /** 2 * 3 */ 4 package solution; 5 6 import java.util.HashMap; 7 8 /** 9 * @author whh 10 * 11 * Given a string S and a string T, find the minimum window in S which 12 * will contain all the characters in T in complexity O(n). 13 * 14 * For example, S = "ADOBECODEBANC", T = "ABC" Minimum window is "BANC". 15 * 16 * Note: 17 * 18 * If there is no such window in S that covers all characters in T, 19 * return the emtpy string "". If there are multiple such windows, you 20 * are guaranteed that there will always be only one unique minimum 21 * window in S. 22 */ 23 public class MinimumWindowSubstring { 24 25 /** 26 * @param args 27 */ 28 public static void main(String[] args) { 29 MinimumWindowSubstring mws = new MinimumWindowSubstring(); 30 String S = "aabbDEE", T = "aabbDEE"; 31 System.out.println(mws.minWindow(S, T)); 32 } 33 34 /** 35 * @param S 36 * @param T 37 * @return 38 */ 39 public String minWindow(String S, String T) { 40 41 HashMap<Character, Integer> hasFound = new HashMap<Character, Integer>(); 42 HashMap<Character, Integer> needToFind = new HashMap<Character, Integer>(); 43 44 for (int i = 0; i < T.length(); i++) { 45 hasFound.put(T.charAt(i), 0); 46 47 if (needToFind.containsKey(T.charAt(i))) { 48 needToFind.put(T.charAt(i), needToFind.get(T.charAt(i)) + 1); 49 } else { 50 needToFind.put(T.charAt(i), 1); 51 } 52 } 53 54 int begin = 0; 55 int minWindowSize = S.length(); 56 String retString = ""; 57 58 int count = 0; 59 60 for (int end = 0; end < S.length(); end++) { 61 Character end_c = S.charAt(end); 62 if (needToFind.containsKey(end_c)) { 63 hasFound.put(end_c, hasFound.get(end_c) + 1); 64 if (hasFound.get(end_c) <= needToFind.get(end_c)) { 65 count++; 66 } 67 if (count == T.length()) { 68 while ((!needToFind.containsKey(S.charAt(begin))) 69 || (hasFound.get(S.charAt(begin)) > needToFind 70 .get(S.charAt(begin)))) { 71 72 if (needToFind.containsKey(S.charAt(begin))) { 73 hasFound.put(S.charAt(begin), 74 hasFound.get(S.charAt(begin)) - 1); 75 } 76 77 begin++; 78 } 79 80 if ((end - begin + 1) <= minWindowSize) { 81 minWindowSize = end - begin + 1; 82 retString = S.substring(begin, end + 1); 83 } 84 } 85 } 86 } 87 88 return retString; 89 } 90 91 }
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