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  • Intersection of Two Arrays II

    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?

    解法一:hashmap

    public class Solution {
        public int[] intersect(int[] nums1, int[] nums2) {
            HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
            ArrayList<Integer> result = new ArrayList<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);
            }
        
            for(int i = 0; i < nums2.length; i++)
            {
                if(map.containsKey(nums2[i]) && map.get(nums2[i]) > 0)
                {
                    result.add(nums2[i]);
                    map.put(nums2[i], map.get(nums2[i])-1);
                }
            }
        
           int[] r = new int[result.size()];
           for(int i = 0; i < result.size(); i++)
           {
               r[i] = result.get(i);
           }
        
           return r;
        }
    }

    解法二:

    Sort both arrays, use two pointers

    Time complexity: O(nlogn)

    public class Solution {
        public int[] intersect(int[] nums1, int[] nums2) {
            List<Integer> res = new ArrayList<Integer>();
            Arrays.sort(nums1);
            Arrays.sort(nums2);
            for(int i = 0, j = 0; i< nums1.length && j<nums2.length;){
                    if(nums1[i]<nums2[j]){
                        i++;
                    }
                    else if(nums1[i] == nums2[j]){
                        res.add(nums1[i]);
                        i++;
                        j++;
                    }
                    else{
                        j++;
                    }
            }
            int[] result = new int[res.size()];
            for(int i = 0; i<res.size();i++){
                result[i] = res.get(i);
            }
            return result;
        }
    }
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  • 原文地址:https://www.cnblogs.com/hygeia/p/5700021.html
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