Algorithm questions in an Alibaba telephone interview

Telephone Interview algorithm question: Find the elements with the most repeated entries in the array and print

 

The problem is not difficult. You can give a better solution.

 

 

Java code

1. Import java. util. hashmap;
2. Import java. util. iterator;
3. Import java. util. Map. entry;
5. Import commons. algorithm. Sort. quicksort;
7. /**
8. * Find the most repeated elements in the array and print them.
9. *
10. */
11. Public class problem_3 {
13. // Search for O (N * log2 (n) + n) cyclically after fast sorting)
14. Public static void find1 (INT [] ARR ){
16. Quicksort. Sort (ARR );
19. Int max = arr [0];
20. Int pre = 1;
21. Int now = 1;
23. For (INT I = 0; I <(ARR. Length-1); I ++ ){
25. If (ARR [I] = arr [I + 1])
26. Now ++;
27. Else {
28. If (now> = pre ){
29. Pre = now;
30. Now = 1;
31. Max = arr [I];
32. }
33. }
34. }
38. }
40. // Nested loop query O (N * n)
41. Public static void find2 (INT [] ARR ){
44. Int pre = 0;
45. Int max = arr [0];
47. For (INT I = 0; I <arr. length; I ++ ){
48. Int now = 0;
49. For (Int J = 0; j <arr. length; j ++ ){
51. If (ARR [I] = arr [J]) {
52. Now ++;
53. }
54. }
56. If (now> = pre ){
57. Max = arr [I];
58. Pre = now;
59. }
62. }
66. }
68. // Hash
69. Public static void find3 (INT [] ARR ){
70. Hashmap <integer, integer> Hm = new hashmap <integer, integer> ();
71. For (INT I = 0; I <arr. length; I ++ ){
72. If (Hm. containskey (ARR [I]) {
73. Int COUNT = hm. Get (ARR [I]);
74. Hm. Put (ARR [I], ++ count );
76. } Else {
77. Hm. Put (ARR [I], 1 );
78. }
79. }
81. Iterator <entry <integer, integer> it = hm. entryset (). iterator ();
82. Int pre = 0;
83. Int max = arr [0];
84. While (it. hasnext ()){
85. Entry <integer, integer> en = it. Next ();
86. Int key = en. getkey ();
87. Int val = en. getvalue ();
88. If (Val> pre ){
89. Pre = val;
90. Max = key;
91. }
93. }
97. }
98. Public static void main (string ARGs []) {
100. // There are more than 800 duplicate elements in the amount of data, which are: 46 3680 195
101. Int arr2 [] = {0, 1, 2 ,.....
102. , 0, 1, 2, 3, 6, 7, 8, 9 };
104. // The amount of data is 800. There are few duplicate elements, and the search time is 82 3727 360.
105. Int arr [] = {, 6 ......
106. , 730,731, 52, 53, 3,794, 95,796,797,798,799 };
110. Long start, end;
112. Start = system. currenttimemillis ();
113. For (INT I = 0; I <1000; I ++) find1 (ARR );
114. End = system. currenttimemillis ();
115. System. Out. println (end-Start );
117. Start = system. currenttimemillis ();
118. For (INT I = 0; I <1000; I ++) find2 (ARR );
119. End = system. currenttimemillis ();
120. System. Out. println (end-Start );
123. Start = system. currenttimemillis ();
124. For (INT I = 0; I <1000; I ++) find3 (ARR );
125. End = system. currenttimemillis ();
126. System. Out. println (end-Start );
128. }
129. }

Reprinted from: http://leves.javaeye.com/blog/777508