leetcode 349：两个数组的交集I

Problem:

Given two arrays, write a function to compute their intersection.

中文：已知两个数组，写一个函数来计算它们的交集

Example:

• Given nums1 = [1, 2, 2, 1], nums2 = [2, 2],return [2, 2].

  ---

• 已知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.

  ---

• （每个元素出现的次数与原来两个数组中重复的次数相同）

• （数组的元素可以是任意顺序的）

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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• 假如已知的数组是有序的该如何优化算法？
• 假如数组1的大小比数组2要小该如何优化？
• 假如数组2的元素在磁盘上是排序好的，但是内存不能一次性容纳这些元素，该怎么做？

Solution:

Analysis:

• 1.直接的想法：直接嵌套遍历两个数组，遇到相同元素的就加入一个预定义好的数组中。
  
  • 预定义的数组长度是数组1和数组2中较小的一个。
  • 最后将该数组有效的元素重新移植到新的一个数组中。
• 2.修正1：得到的数组可能有很多重复的元素。
  
  • 要再添加新元素到数组中时检查数组中是否已经存在该元素。
  • 数组会越界，所以要在添加前检查元素个数是否超过了数组长度。

Code in JAVA

public static int[] intersection(int[] a1, int[] a2) {
       int n = Math.min(a1.length, a2.length);
       int[] is = new int[n];
       int count = 0;
       for(int i = 0; i < a1.length; i++){
        int tmep = a1[i];
        for(int j = 0; j < a2.length; j++){
            if(tmep == a2[j]){
                boolean exist = false;
                for(int k = 0; k < count; k++){
                    if(is[k] == tmep){
                        exist = true;
                        break;
                    }
                }
                if(count >= n){
                    break;
                }
                if(!exist){
                    is[count] = tmep;
                    count++;
                }

                break;
            }
        }
       }
       int[] itersection = new int[count];
       for(int i = 0; i < count; i++){
        itersection[i] = is[i];
       }

       return itersection;
}