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LeetCode:Intersection of Two Arrays II

时间:2016-05-22 12:14:20      阅读:153      评论:0      收藏:0      [点我收藏+]

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Intersection of Two Arrays II




Total Accepted: 1340 Total Submissions: 3078 Difficulty: Easy

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 num2‘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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java code:

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        
        int len1 = nums2.length,len2 = nums2.length;
        
        List<Integer> list = new ArrayList<Integer>();
        List<Integer> interList = new ArrayList<Integer>();
        
        for(int num:nums1) {
            list.add(num);
        }
        
        for(int i=0;i<len2;i++) {
            if(list.contains(nums2[i])) {
                interList.add(nums2[i]);
                list.remove(list.indexOf(nums2[i]));
            }
        }
        
        int[] ret = new int[interList.size()];
        int cnt = 0;
        for(int num:interList) {
            ret[cnt++] = num;
        }
        
        return ret;
    }
}


LeetCode:Intersection of Two Arrays II

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原文地址:http://blog.csdn.net/itismelzp/article/details/51474073

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