Intersection of the two arrays

arai60/intersection_of_the_two_arrays/phase1.py

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+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        nums1 = set(nums1)
+        nums2 = set(nums2)
+        return list(nums1 & nums2)

arai60/intersection_of_the_two_arrays/phase2.py

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+"""
+Reference
+ryotaro: https://github.com/Ryotaro25/leetcode_first60/pull/14/files
+mike: https://github.com/Mike0121/LeetCode/pull/30/files
+kazukiii: https://github.com/kazukiii/leetcode/pull/14/files
+本題ではないが, pythonのsortedとlist.sort()の計算中に空間計算量O(n)を消費する情報を得られてラッキー
+
+numpy実装も確認したが,,,python側から見えるのはnp.arrayが実装されている前提のものだった
+https://github.com/numpy/numpy/blob/v2.0.0/numpy/lib/_arraysetops_impl.py#L578-L668
+"""
+
+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        return list(set(nums1) & set(nums2))
+
+# sortを用いて
+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        nums1 = sorted(nums1)
+        nums2 = sorted(nums2)
+        intersection_of_the_two = []
+        seen = set()
+        i1 = 0
+        i2 = 0
+        while i1 < len(nums1) and i2 < len(nums2):
+            if nums1[i1] == nums2[i2]:
+                if nums1[i1] not in seen:
+                    intersection_of_the_two.append(nums1[i1])
+                    seen.add(nums1[i1])
+                i1 += 1
+                i2 += 1
+            elif nums1[i1] < nums2[i2]:
+                i1 += 1
+            else:
+                i2 += 1
+        return intersection_of_the_two

arai60/intersection_of_the_two_arrays/phase3.py

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+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        return list(set(nums1) & set(nums2))