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944天前下面这个方法 要求两个数组是已排序的 |
Java codepackage com.haojia.algorithm; |
/** |
* 求两个已排序的数组的交集 |
* |
* @author July |
* |
*/ |
public class Intersect { |
public static void go(int[] a, int[] b) { |
int i = 0; |
int j = 0; |
while (i < a.length && j < b.length) { |
if (a[i] < b[j]) { |
i++; |
} else if (a[i] > b[j]) { |
j++; |
} else { |
System.out.print(a[i] + " "); |
i++; |
j++; |
} |
} |
} |
public static void main(String[] args) { |
int[] a = { 1, 5, 8, 10, 14, 15, 17, 18, 20, 22, 24, 25, 28 }; |
int[] b = { 2, 4, 6, 8, 10, 12 }; |
go(a, b); |
} |
} |
------解决方案-------------------- |
Java codeimport java.util.Arrays; |
import java.util.HashSet; |
import java.util.Random; |
import java.util.Set; |
public class SelectBag { |
public static int find(int key,int[] N){ |
int lb=0; |
int ub=N.length-1; |
int in; |
while(true){ |
in=(lb+ub)/2; |
if(N[in]==key) |
return in; |
else if(lb>ub) |
return -1; |
else{ |
if(N[in]<key) |
lb=in+1; |
else |
ub=in-1; |
} |
} |
} |
public static void main(String[] args){ |
int[] M=RandomIntArray(30); |
int[] N=RandomIntArray(30); |
System.out.println(Arrays.toString(M)); |
System.out.println(Arrays.toString(N)); |
Set<Integer> list=new HashSet<Integer>(); |
for(int m:M){ |
int getInt=find(m,N); |
if(getInt!=-1) |
list.add(N[getInt]); |
} |
System.out.println("两个数组中重复的元素有："+list); |
} |
|
public static int[] RandomIntArray(int count){ |
int[] array=new int[count]; |
Random r=new Random(); |
for(int i=0;i<count;i++){ |
array[i]=r.nextInt(100); |
} |
Arrays.sort(array); |
return array; |
} |
|
} |
结果： |
[1, 3, 8, 11, 12, 20, 22, 22, 31, 34, 37, 40, 41, 45, 48, 49, 50, 51, 57, 69, 72, 73, 84, 84, 85, 93, 93, 93, 98, 98] |
[0, 4, 4, 9, 12, 16, 26, 26, 28, 32, 36, 41, 42, 44, 48, 48, 55, 56, 61, 65, 72, 72, 73, 75, 76, 78, 83, 84, 89, 94] |
两个数组中重复的元素有：[84, 48, 41, 72, 12, 73]//源代码片段来自云代码http://yuncode.net |
|
