commit solution 350

Author: xiao2shiqi

README.md

@@ -20,6 +20,7 @@ LeetCode
 |509|[Fibonacci Number](https://leetcode-cn.com/problems/fibonacci-number/) | [Java](https://github.com/xiao2shiqi/leetcode/blob/master/src/main/java/S509.java), [Ruby](https://github.com/xiao2shiqi/leetcode/blob/master/ruby/509.rb) | Easy
 |448|[Find All Numbers Disappeared in an Array](https://leetcode-cn.com/problems/find-all-numbers-disappeared-in-an-array/) | [Java](https://github.com/xiao2shiqi/leetcode/blob/master/src/main/java/S448.java), [Ruby](https://github.com/xiao2shiqi/leetcode/blob/master/ruby/448.rb) | Easy
 |414|[Third Maximum Number](https://leetcode-cn.com/problems/third-maximum-number/) | [Java](https://github.com/xiao2shiqi/leetcode/blob/master/src/main/java/S414.java) | Easy
+|350|[Intersection of Two Arrays II](https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/) | [Java](https://github.com/xiao2shiqi/leetcode/blob/master/src/main/java/S350.java) | Easy
 |283|[Move Zeroes](https://leetcode-cn.com/problems/move-zeroes/) | [Java](https://github.com/xiao2shiqi/leetcode/blob/master/src/main/java/S283.java), [Ruby](https://github.com/xiao2shiqi/leetcode/blob/master/ruby/283.rb) | Easy
 |268|[Missing Number](https://leetcode-cn.com/problems/missing-number/) | [Java](https://github.com/xiao2shiqi/leetcode/blob/master/src/main/java/S268.java), [Ruby](https://github.com/xiao2shiqi/leetcode/blob/master/ruby/268.rb) | Easy
 |263|[Ugly Number](https://leetcode-cn.com/problems/ugly-number/) | [Java](https://github.com/xiao2shiqi/leetcode/blob/master/src/main/java/S263.java), [Go](https://github.com/xiao2shiqi/leetcode/blob/master/go/263.go) | Easy

src/main/java/S350.java

@@ -0,0 +1,41 @@
+import java.util.*;
+
+/**
+ * LC#350： 两个数组的交集 II
+ * Link：https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/
+ * 思路1：将数组放进 hash 桶中记数，取出当 count > 1 则为数组的交集，最终重定义数组范围
+ * 思路2：排序 + 双指针遍历，遍历两个数组，如果元素相等则加入结果集，如果元素不相等则较小的元素右移一位，直到数组遍历结束，时间复杂度 log n，空间复杂度O(1)
+ * 思路3：直接使用 Java List 容器实现 （时间，空间效率都很低）
+ * @author Phoenix on 2021/6/26.
+ */
+public class S350 {
+
+    public int[] intersect(int[] nums1, int[] nums2) {
+        List<Integer> list1 = new ArrayList<>();
+        for (int num : nums1) {
+            list1.add(num);
+        }
+        List<Integer> list2 = new ArrayList<>();
+        for (int num : nums2) {
+            if (list1.contains(num)) {
+                list2.add(num);
+                // 从 list1 除去已匹配的数值
+                list1.remove(Integer.valueOf(num));
+            }
+        }
+        int[] res = new int[list2.size()];
+        int i = 0;
+        for (int num : list2) {
+            res[i++] = num;
+        }
+        return res;
+    }
+
+    public static void main(String[] args) {
+        int[] nums1 = {1, 2, 2, 1};
+        int[] nums2 = {2, 2};
+
+        int[] ints = new S350().intersect(nums1, nums2);
+        System.out.println(Arrays.toString(ints));
+    }
+}