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?
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題目
計(jì)算兩個(gè)數(shù)組的交
樣例
nums1 = [1, 2, 2, 1], nums2 = [2, 2], 返回 [2, 2].
分析
這道題是上一道題的擴(kuò)展，只是這次要記錄重復(fù)的元素的個(gè)數(shù)，這次我們就用一個(gè)哈希表申眼，鍵記錄重復(fù)元素烤黍，值記錄重復(fù)個(gè)數(shù)就行了败去。采用hashset的方法
代碼
public class Solution {
/**
* @param nums1 an integer array
* @param nums2 an integer array
* @return an integer array
*/
public int[] intersection(int[] nums1, int[] nums2) {
// Write your code here
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i < nums1.length; ++i) {
if (map.containsKey(nums1[i]))
map.put(nums1[i], map.get(nums1[i]) + 1);
else
map.put(nums1[i], 1);
}
List<Integer> results = new ArrayList<Integer>();
for (int i = 0; i < nums2.length; ++i)
if (map.containsKey(nums2[i]) &&
map.get(nums2[i]) > 0) {
results.add(nums2[i]);
map.put(nums2[i], map.get(nums2[i]) - 1);
}
int result[] = new int[results.size()];
for(int i = 0; i < results.size(); ++i)
result[i] = results.get(i);
return result;
}
}
