求两个有序整数集合的交集，比比谁的算法快！
阿里巴巴一道笔试题题目：有两个有序整数集合a和b，写一个函数找出它们的交集？
方法一：
1 private static Set<Integer> setMethod(int[] a,int[] b){
2 Set<Integer> set = new HashSet<Integer>();
3 Set<Integer> set2 = new HashSet<Integer>();
4 for(int i=0; i<a.length; i++) {
5 set.add(a[i]);
6 }
7 for(int j=0; j<b.length; j++) {
8 if(!set.add(b[j]))
9 set2.add(b[j]);
10 }
11 return set2;
12 }
方法二：
1 private static Set<Integer> forMethod(int[] a,int[] b){
2 Set<Integer> set=new HashSet<Integer>();
3 int i=0,j=0;
4 while(i<a.length && j<b.length){
5 if(a[i]<b[j])
6 i++;
7 else if(a[i]>b[j])
8 j++;
9 else{
10 set.add(a[i]);
11 i++;
12 j++;
13 }
14 }
15 return set;
16 }
方法三：
1 private static int[] intersect(int[] a, int[] b) {
2 if (a[0] > b[b.length - 1] || b[0] > a[a.length - 1]) {
3 return new int[0];
4 }
5 int[] intersection = new int[Math.max(a.length, b.length)];
6 int offset = 0;
7 for (int i = 0, s = i; i < a.length && s < b.length; i++) {
8 while (a[i] > b[s]) {
9 s++;
10 }
11 if (a[i] == b[s]) {
12 intersection[offset++] = b[s++];
13 }
14 while (i < (a.length - 1) && a[i] == a[i + 1]) {
15 i++;
16 }
17 }
18 if (intersection.length == offset) {
19 return intersection;
20 }
21 int[] duplicate = new int[offset];
22 System.arraycopy(intersection, 0, duplicate, 0, offset);
23 return duplicate;
24 }
三种方法的性能对比测试：
1 public class NumberCrossTest {
2 public static void main(String[] args) {
3 int[] a1 = new int[100000];
4 for (int i = 0; i < a1.length; i++) {
5 a1[i] = i + 10;
6 }
7 int[] a2 = new int[200000];
8 for (int i = 0; i < a2.length; i++) {
9 a2[i] = i + 20;
10 }
11 long begin = System.currentTimeMillis();
12 Set<Integer> set1 = setMethod(a1, a2);
13 long end = System.currentTimeMillis();
14 System.out.println(end - begin);// 359
15 begin = System.currentTimeMillis();
16 Set<Integer> set2 = forMethod(a1, a2);
17 end = System.currentTimeMillis();
18 System.out.println(end - begin);// 160
19 begin = System.currentTimeMillis();
20 int[] c = intersect(a1, a2);
21 end = System.currentTimeMillis();
22 System.out.println(end - begin);// 10
23 // 测试两种方法的结果是否相等
24 System.out.println(set1.equals(set2));// true
25 Set<Integer> set3 = new HashSet<Integer>();
26 for (int i = 0; i < c.length; i++) {
27 set3.add(c[i]);
28 }
29 System.out.println(set1.equals(set3));// true
30 }
31
32 private static Set<Integer> setMethod(int[] a, int[] b) {
33 Set<Integer> set = new HashSet<Integer>();
34 Set<Integer> set2 = new HashSet<Integer>();
35 for (int i = 0; i < a.length; i++) {
36 set.add(a[i]);
37 }
38 for (int j = 0; j < b.length; j++) {
39 if (!set.add(b[j]))
40 set2.add(b[j]);
41 }
42 return set2;
43 }
44
45 private static Set<Integer> forMethod(int[] a, int[] b) {
46 Set<Integer> set = new HashSet<Integer>();
47 int i = 0, j = 0;
48 while (i < a.length && j < b.length) {
49 if (a[i] < b[j])
50 i++;
51 else if (a[i] > b[j])
52 j++;
53 else {
54 set.add(a[i]);
55 i++;
56 j++;
57 }
58 }
59 return set;
60 }
61
62 private static int[] intersect(int[] a, int[] b) {
63 if (a[0] > b[b.length - 1] || b[0] > a[a.length - 1]) {
64 return new int[0];
65 }
66 int[] intersection = new int[Math.max(a.length, b.length)];
67 int offset = 0;
68 for (int i = 0, s = i; i < a.length && s < b.length; i++) {
69 while (a[i] > b[s]) {
70 s++;
71 }
72 if (a[i] == b[s]) {
73 intersection[offset++] = b[s++];
74 }
75 while (i < (a.length - 1) && a[i] == a[i + 1]) {
76 i++;
77 }
78 }
79 if (intersection.length == offset) {
80 return intersection;
81 }
82 int[] duplicate = new int[offset];
83 System.arraycopy(intersection, 0, duplicate, 0, offset);
84 return duplicate;
85 }
86 }
结果对比：
方法一用时：359 毫秒
方法二用时：160 毫秒
方法三用时：10 毫秒
