leetcode -- Substring with Concatenation of All Words

 思路：

从左往右扫描主字符串，用两个map来统计字符串的出现次数

 这里L数组中可能会出现两个相同的字符串

 1 import java.util.Collection;
 2 public class Solution {
 3     public ArrayList<Integer> findSubstring(String S, String[] L) {
 4          ArrayList<Integer> results = new ArrayList<Integer>();
 5         int len = L.length;
 6         if (len == 0) {
 7             results.add(0);
 8             return results;
 9         }
10 
11         int M = S.length();
12         int N = L[0].length();
13         Map<String, Integer> expected = new HashMap<String, Integer>();
14         Map<String, Integer> real = new HashMap<String, Integer>();
15         for (int j = 0; j < len; j++) {
16             if(expected.get(L[j]) == null){
17                 expected.put(L[j], 1);
18                 real.put(L[j], 0);
19             } else{
20                 expected.put(L[j], expected.get(L[j]) + 1);
21             }
22         }
23 
24         for (int i = 0; i <= (M - N * len);) {
25             int j = i;
26             for (; j < i + N * len;) {
27                 String tmp = S.substring(j, j + N);
28                 if(expected.get(tmp) == null){
29                     i = i + 1;
30                     for (int m = 0; m < len; m++) {
31                         real.put(L[m], 0);
32                     }
33                     break;
34                 }else {
35                     real.put(tmp, real.get(tmp) + 1);
36                     j = j + N;
37                 } 
38             }
39 
40             if (j >= i + N * len) {
41                 boolean flag = true;
42                 for(int n = 0; n < L.length; n++){
43                     if(expected.get(L[n]) != real.get(L[n])){
44                         flag = false;
45                         break;
46                     }
47                 }
48                 
49                 if(flag){
50                     results.add(i);
51                     i = i + N;
52                 } else {
53                     i = i + 1;
54                 }
55 
56                 for (int m = 0; m < len; m++) {
57                     real.put(L[m], 0);
58                 }
59             }
60         }
61         return results;
62     }
63 }

 

 

重构后代码：

 1 public static ArrayList<Integer> findSubstring3(String S, String[] L) {
 2         ArrayList<Integer> results = new ArrayList<Integer>();
 3         int len = L.length;
 4         if (len == 0) {
 5             results.add(0);
 6             return results;
 7         }
 8 
 9         int M = S.length();
10         int N = L[0].length();
11         Map<String, Integer> expected = new HashMap<String, Integer>();
12         Map<String, Integer> real = new HashMap<String, Integer>();
13         for (int j = 0; j < len; j++) {
14             if (expected.get(L[j]) == null) {
15                 expected.put(L[j], 1);
16                 real.put(L[j], 0);
17             } else {
18                 expected.put(L[j], expected.get(L[j]) + 1);
19             }
20         }
21 
22         for (int i = 0; i <= (M - N * len); i++) {
23             int j = 0;
24             for (; j < len;) {
25                 String tmp = S.substring(i + j * N, i + (j + 1) * N);
26                 if (expected.get(tmp) == null) {
27                     break;
28                 } else {
29                     real.put(tmp, real.get(tmp) + 1);
30                     if (real.get(tmp) > expected.get(tmp)) {
31                         break;
32                     }
33                     j = j + 1;
34                 }
35             }
36 
37             if (j == len) {
38                 results.add(i);
39             }
40             for (int m = 0; m < len; m++) {
41                 real.put(L[m], 0);
42             }
43 
44         }
45         return results;
46     }

易错的几个数据集：

String S = "lingmindraboofooowingdingbarrwingmonkeypoundcake";
String[] L = new String[] { "fooo", "barr", "wing", "ding", "wing" };

 

String S2 = "aaa";
String[] L2 = new String[] { "a", "b" };

 

String S3 = "barfoothefoobarman";
String[] L3 = new String[] { "foo", "bar" };