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[英]How to use HashSet to find common elements in two Comparable arrays?
[英]Find all common elements of two sorted arrays use binary search
我已经开始学习Java,所以我有一些愚蠢的问题。 我想找到两个排序的 arrays 的所有共同元素,使用二进制搜索,下面是我的代码。 它运行没有正确的结果。
import java.util.Arrays;
public class CommonElements{
public static void commonElements(int[] iArr1, int[] iArr2) {
Arrays.sort(iArr1);
Arrays.sort(iArr2);
int index2 = 0, index1 = 0;
int left1 = 0, right1 = (iArr1.length)-1;
while (index2 < iArr2.length && index1 < iArr1.length) {
while (left1 <= right1) {
int mid1 = (left1 + right1)/2;
if (iArr2[index2] == iArr1[mid1]) {
System.out.print(iArr1[mid1] + " ");
index2++;
} else if (iArr2[index2] > iArr1[mid1])
left1 = mid1 + 1;
else right1 = mid1 - 1;
}
}
}
public static void main(String[] args) {
int[] iArr1 = {6,-3,9,-1,0,0,8};
int[] iArr2 = {-7,6,-3,-5};
commonElements(iArr1,iArr2);
System.out.println("END");
}
}
当您可以使用两个指针在两个已排序的 arrays 中查找公共元素时,为什么要对数组进行二进制搜索? 您可以在每次迭代中递增指向两个指针中较小值的指针,如果两个值相等,则表示该元素在两个 arrays 中是公共的,现在递增两个指针并打印公共元素。
public static void commonElements(int[] iArr1, int[] iArr2) {
Arrays.sort(iArr1);
Arrays.sort(iArr2);
int index2 = 0, index1 = 0;
while (index2 < iArr2.length && index1 < iArr1.length) {
if (iArr2[index2] < iArr1[index1]) {
index2++;
} else if (iArr2[index2] > iArr1[index1]) {
index1++;
} else {
System.out.print(iArr1[index1] + " ");
index1++;
index2++;
}
}
}
输入:
iArr1 => [-3,-1,0,0,6,8,9], iArr2 => [-7,-5,-3,6]
Output: -3 6
如果需要二进制搜索,那么您可以将代码编辑为这样的内容,以使其不会陷入无限循环(我已使用注释标记了更改):
public static void commonElements(int[] iArr1, int[] iArr2) {
Arrays.sort(iArr1);
Arrays.sort(iArr2);
int index2 = 0, index1 = 0;
while (index2 < iArr2.length && index1 < iArr1.length) {
int left1 = 0, right1 = (iArr1.length)-1; // 1) Reinitialize left1 and right1 in each iteration of iArr2.
while (left1 <= right1) {
int mid1 = (left1 + right1)/2;
if (iArr2[index2] == iArr1[mid1]) {
System.out.print(iArr1[mid1] + " ");
break; // 2) Break out of the binary search when you've found the common element.
} else if (iArr2[index2] > iArr1[mid1])
left1 = mid1 + 1;
else right1 = mid1 - 1;
}
index2++; // 3) Move to next item of iArr2.
}
}
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