add 350

Author: andavid

README.md

@@ -21,6 +21,7 @@
 | 189  | [Rotate Array][189]                         |
 | 217  | [Contains Duplicate][217]                   |
 | 136  | [Single Number][136]                        |
+| 350  | [Intersection of Two Arrays II][350]        |
 
 [leetcode]: https://leetcode.com/problemset/all/
 [blankj]: https://github.com/Blankj/awesome-java-leetcode
@@ -32,3 +33,4 @@
 [189]: https://github.com/andavid/leetcode-java/blob/master/note/189/README.md
 [217]: https://github.com/andavid/leetcode-java/blob/master/note/217/README.md
 [136]: https://github.com/andavid/leetcode-java/blob/master/note/136/README.md
+[350]: https://github.com/andavid/leetcode-java/blob/master/note/350/README.md

note/350/README.md

@@ -0,0 +1,62 @@
+# [Intersection of Two Arrays II][title]
+
+## Description
+
+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 nums2'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?
+
+## 思路
+使用 HashMap 遍历第一个数组，并记录数组元素出现的次数。
+然后遍历第二个数组，如果数组元素在 HashMap 中，则把该元素添加到 List，同时 HashMap 该元素出现次数减一。
+最后把 List 转成 int 数组即可。
+时间复杂度 O(n)，空间复杂度 O(n)。
+
+## [完整代码][src]
+
+```java
+class Solution {
+  public int[] intersect(int[] nums1, int[] nums2) {
+    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
+    List<Integer> list = new ArrayList<Integer>();
+
+    for (int num : nums1) {
+      if (map.containsKey(num)) {
+        map.put(num, map.get(num) + 1);
+      } else {
+        map.put(num, 1);
+      }
+    }
+
+    for (int num : nums2) {
+      if (map.containsKey(num)) {
+        if (map.get(num) > 0) {
+          list.add(num);
+          map.put(num, map.get(num) - 1);
+        }
+      }
+    }
+
+    int size = list.size();
+    int result[] = new int[size];
+    for (int i = 0; i < size; i++) {
+      result[i] = list.get(i);
+    }
+
+    return result;
+  }
+}
+```
+
+[title]: https://leetcode.com/problems/intersection-of-two-arrays-ii
+[src]: https://github.com/andavid/leetcode-java/blob/master/src/com/andavid/leetcode/_350/Solution.java

src/com/andavid/leetcode/_350/Solution.java

@@ -0,0 +1,41 @@
+import java.util.*;
+
+class Solution {
+  public int[] intersect(int[] nums1, int[] nums2) {
+    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
+    List<Integer> list = new ArrayList<Integer>();
+
+    for (int num : nums1) {
+      if (map.containsKey(num)) {
+        map.put(num, map.get(num) + 1);
+      } else {
+        map.put(num, 1);
+      }
+    }
+
+    for (int num : nums2) {
+      if (map.containsKey(num)) {
+        if (map.get(num) > 0) {
+          list.add(num);
+          map.put(num, map.get(num) - 1);
+        }
+      }
+    }
+
+    int size = list.size();
+    int result[] = new int[size];
+    for (int i = 0; i < size; i++) {
+      result[i] = list.get(i);
+    }
+
+    return result;
+  }
+
+  public static void main(String[] args) {
+    Solution solution = new Solution();
+    int[] nums = solution.intersect(new int[]{1, 2, 2, 1}, new int[]{2, 2});
+    for (int i = 0; i < nums.length; i++) {
+      System.out.print(i == 0 ? nums[i] : "," + nums[i]);
+    }
+  }
+}