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;

}