LeetCode 350. Intersection of Two Arrays II
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2022-07-15 10:45:55
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题目描述
- Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2]. - Note:
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order. - Follow up:
What if the given array is already sorted? How would you optimize your algorithm?
What if nums1’s size is small compared to nums2’s size? Which algorithm is better?
What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once? - 地址
问题分析
- 找两个数组的交集,重复也计入
- 用HashMap统计nums1数组中的字符及其相应次数,然后遍历nums2看是否在map中并且次数大于0,若在,加入结果集,并减小相应次数(若是不计重复,则可用HashSet ,如LeetCode 349)
- 排序后用双指针
- 排序后用二分
- 对于附加问题,这才是重点:附上discuss区答案
代码实现
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
if (nums1 == null || nums2 == null || nums1.length == 0 || nums2.length == 0) {
return new int[0];
}
Arrays.sort(nums1);
Arrays.sort(nums2);
int i = 0;
int j = 0;
ArrayList<Integer> list = new ArrayList<>();
while (i < nums1.length && j < nums2.length) {
if (nums1[i] > nums2[j]) {
j++;
}else if (nums1[i] < nums2[j]) {
i++;
}else {
list.add(nums1[i]);
i++;
j++;
}
}
int[] res = new int[list.size()];
for (int k = 0; k < list.size(); k++) {
res[k] = list.get(k);
}
return res;
}
/*
public int[] intersect(int[] nums1, int[] nums2) {
if (nums1 == null || nums2 == null || nums1.length == 0 || nums2.length == 0) {
return new int[0];
}
HashMap<Integer,Integer> map = new HashMap<>();
for (int num : nums1) {
if (map.containsKey(num)) {
map.put(num, map.get(num) + 1);
}else {
map.put(num, 1);
}
}
ArrayList<Integer> list = new ArrayList<>();
for (int num : nums2) {
if (map.containsKey(num) && map.get(num) > 0) {
list.add(num);
map.put(num, map.get(num) - 1);
}
}
int[] res = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
res[i] = list.get(i);
}
return res;
}
*/
}
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