intersection of two arra I_II easy

Author: weiy

Array/IntersectionOfTwoArrays.py

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+"""
+Given two arrays, write a function to compute their intersection.
+
+Example 1:
+
+Input: nums1 = [1,2,2,1], nums2 = [2,2]
+Output: [2]
+Example 2:
+
+Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
+Output: [9,4]
+Note:
+
+Each element in the result must be unique.
+The result can be in any order.
+
+
+找两个数组里的重复数据。
+
+唯一，任意顺序。
+
+集合，取交集即可。
+
+零启动。
+
+beat 100% 20ms.
+
+测试地址：
+https://leetcode.com/problems/intersection-of-two-arrays/description/
+
+"""
+class Solution(object):
+    def intersection(self, nums1, nums2):
+        """
+        :type nums1: List[int]
+        :type nums2: List[int]
+        :rtype: List[int]
+        """
+        
+        return list(set(nums1) & set(nums2))

Array/IntersectionOfTwoArraysII.py

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+"""
+Given two arrays, write a function to compute their intersection.
+
+Example 1:
+
+Input: nums1 = [1,2,2,1], nums2 = [2,2]
+Output: [2,2]
+Example 2:
+
+Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
+Output: [4,9]
+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?
+
+与 1 差不多，不过这次有重复。
+进阶条件里问：
+1. 如果是已排序的可以怎么优化？
+2. 如果 数组1 比 数组2 要小，哪个算法好些？
+3. 如果 数组2 在硬盘里排过序，但内存不足以一次性全部读取该怎么做？
+
+我的思路是：
+1. 先排序。
+2. 之后设置两个指针。
+
+若 1与2相同，则将结果添加到最终结果的列表中，1和2各+1。
+若 1比2要大，那么表示2向后走还有可能遇到和1相同的数字，所以2 +1。
+否则 1 +1。直到有一个到了末尾。
+
+这个思路的话，进阶的1和3直接可以包含进去。
+2的话，Discuss 里的其他方法基本上是用 哈希表。 1和2哈希，然后取个数较小的交集，这种方法在数组较小的时候要比上面提到的思路快。
+
+排序后的思路：
+
+beat 100% 24ms.
+
+测试地址：
+https://leetcode.com/problems/intersection-of-two-arrays-ii/description/
+
+
+"""
+class Solution(object):
+    def intersect(self, nums1, nums2):
+        """
+        :type nums1: List[int]
+        :type nums2: List[int]
+        :rtype: List[int]
+        """
+        result = []
+    
+    
+        nums1.sort()
+        nums2.sort()
+        
+        _n1 = 0
+        _n2 = 0
+        
+        _n1_length = len(nums1)
+        _n2_length = len(nums2)
+        
+        while _n1 < _n1_length and _n2 < _n2_length:
+            if nums1[_n1] == nums2[_n2]:
+                result.append(nums1[_n1])
+                _n1 += 1
+                _n2 += 1
+            
+            elif nums1[_n1] < nums2[_n2]:
+                _n1 += 1
+            else:
+                _n2 += 1
+        
+        return result