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