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ArrayIntersection.java
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ArrayIntersection.java
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package datastructure.hashtable.leetcode;
import java.util.*;
/**
* @author roseduan
* @time 2020/9/14 4:26 下午
* @description 两个数组的交集
*/
public class ArrayIntersection {
/**
* 使用set的自动去重功能,两个set的交集即是结果
* 运行结果:通过,速度击败26%
*/
public int[] intersection1(int[] nums1, int[] nums2) {
Set<Integer> set1 = new HashSet<>();
for (int i : nums1) {
set1.add(i);
}
Set<Integer> set2 = new HashSet<>();
for (int i : nums2) {
set2.add(i);
}
set1.retainAll(set2);
int[] res = new int[set1.size()];
int k = 0;
for (int i : set1) {
res[k++] = i;
}
return res;
}
/**
* 一个set来解决
* 运行结果:通过,速度击败39%
*/
public int[] intersection2(int[] nums1, int[] nums2) {
Set<Integer> set = new HashSet<>();
for (int i : nums1) {
set.add(i);
}
List<Integer> list = new ArrayList<>();
for (int i : nums2) {
if (set.contains(i)) {
list.add(i);
set.remove(i);
}
}
int[] res = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
res[i] = list.get(i);
}
return res;
}
/**
* 第三种解法:和第二种类似,只是在细节的地方优化了一下
*/
public int[] intersection3(int[] nums1, int[] nums2) {
if (nums1.length > nums2.length) {
return intersection3(nums2, nums1);
}
Set<Integer> set = new HashSet<>();
for (int i : nums1) {
set.add(i);
}
int i = 0;
for (int n : nums2) {
if (set.remove(n)) {
nums1[i++] = n;
}
}
return Arrays.copyOf(nums1, i);
}
}