350. 两个数组的交集 II
2020-02-02 本文已影响0人
Andysys
public int[] intersect(int[] nums1, int[] nums2) {
if (nums1.length == 0 || nums2.length == 0) {
return new int[0];
}
Arrays.sort(nums1);
Arrays.sort(nums2);
List<Integer> list = new ArrayList<>();
int p1 = 0;
int p2 = 0;
while(p1 < nums1.length && p2 < nums2.length) {
if (nums1[p1] < nums2[p2]) {
p1++;
} else if (nums1[p1] > nums2[p2]) {
p2++;
} else {
list.add(nums1[p1]);
p1++;
p2++;
}
}
int[] res = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
res[i] = list.get(i);
}
return res;
}
public int[] intersect2(int[] nums1, int[] nums2) {
if (nums1.length == 0 || nums2.length == 0) {
return new int[0];
}
Map<Integer, Integer> map = new HashMap<>();
List<Integer> list = new ArrayList<>();
for (int num : nums1) {
if (map.containsKey(num)) {
map.put(num, map.get(num) + 1);
} else {
map.put(num, 1);
}
}
for (int num : nums2) {
if (map.containsKey(num)) {
map.put(num, map.get(num) - 1);
if (map.get(num) == 0) {
map.remove(num);
}
list.add(num);
}
}
int size = list.size();
int[] res = new int[size];
for (int i = 0; i < size; i++) {
res[i] = list.get(i);
}
return res;
}