G面经prepare: Set Intersection && Set Difference

 求两个sorted数组的intersection e.g. [1,2,3,4,5],[2,4,6] 结果是[2,4]
difference

类似merge, 分小于等于大于三种情况，然后时间O(m+n), 空间O(1)

package ShortestSubsequenceIncluding;
import java.util.*;

public class Solution2 {
    public ArrayList<Integer> setInters(int[] arr1, int[] arr2) {
        int i1=0, i2=0;
        ArrayList<Integer> res = new ArrayList<Integer>();
        while (i1<arr1.length && i2<arr2.length) {
            if (arr1[i1] < arr2[i2]) i1++;
            else if (arr1[i1] > arr2[i2]) i2++;
            else {
                res.add(arr1[i1]);
                i1++;
                i2++;
            }
        }
        return res;
    }
    
    public ArrayList<Integer> setDiff(int[] arr1, int[] arr2) {
        int i1=0, i2=0;
        ArrayList<Integer> res = new ArrayList<Integer>();
        while (i1<arr1.length && i2<arr2.length) {
            if (arr1[i1] < arr2[i2]) {
                res.add(arr1[i1]);
                i1++;
            }
            else if (arr1[i1] > arr2[i2]) {
                i2++;
            }
            else {
                i1++;
                i2++;
            }
        }
        while (i1 < arr1.length) {
            res.add(arr1[i1]);
            i1++;
        }
        return res;
    }
    

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Solution2 sol = new Solution2();
        ArrayList<Integer> res1 = sol.setInters(new int[]{1,2,3,4,4,5}, new int[]{2,4,4,6});
        ArrayList<Integer> res2 = sol.setDiff(new int[]{2,3,4,4,4,5}, new int[]{1,2,4,4,6});
        System.out.println(res2);

    }

}

如果是无序的两个array，则Build两个hashMap, 遍历第一个map的keySet， O(m+n)time, O(m+n)space