Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].
Note:
- Each element in the result must be unique.
- The result can be in any order.
题目:一目了然找出两个数组的交集
思路：我的大概想法是，先将两个数组分别去重，而后合成一个数组，对该数组进行排序，而后数组中相邻两个元素相等的即为交集元素，即num[i] == num[i-1]
代码有点冗余，效率一般，
public int[] intersection(int[] nums1, int[] nums2) {
int len1 = nums1.length, len2 = nums2.length, j = 0;
if(len1 == 0 || len2 == 0){
return new int[0];
}
HashSet<Integer> set1 = new HashSet<Integer>();
HashSet<Integer> set2 = new HashSet<Integer>();
HashSet<Integer> set3 = new HashSet<Integer>();
for (int i = 0; i < len1; i++) {
set1.add(nums1[i]);
}
for (int i = 0; i < len2; i++) {
set2.add(nums2[i]);
}
int[] temp = new int[set1.size() + set2.size()];
Iterator it1 = set1.iterator();
while (it1.hasNext()) {
temp[j++] = (Integer) it1.next();
}
Iterator it2 = set2.iterator();
while (it2.hasNext()) {
temp[j++] = (Integer) it2.next();
}
Arrays.sort(temp);
for (int i = 1; i < temp.length; i++) {
if (temp[i] == temp[i - 1] ) {
set3.add(temp[i]);
}
}
int res[] = new int[set3.size()];
j = 0;
Iterator it3 = set3.iterator();
while (it3.hasNext()) {
res[j++] = (Integer) it3.next();
}
return res;
}