349. Intersection of Two Arrays --双指针、字典、二分查找、集合

349. Intersection of Two Arrays

很经典的题目，可以使用双指针、字典、二分查找、集合来实现

Given two arrays, write a function to compute their intersection.Example:Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Solution 1:use set operation in python, one-line solution.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))Solution 2:brute-force searching, search each element of the first list in the second list. (to be more efficient, you can sort the second list and use binary search to accelerate)class Solution(object):def intersection(self, nums1, nums2):    """    :type nums1: List[int]    :type nums2: List[int]    :rtype: List[int]    """    res = []    for i in nums1:        if i not in res and i in nums2:            res.append(i)    return resSolution 3:use dict/hashmap to record all nums appeared in the first list, and then check if there are nums in the second list have appeared in the map.class Solution(object):def intersection(self, nums1, nums2):    """    :type nums1: List[int]    :type nums2: List[int]    :rtype: List[int]    """    res = []    map = {}    for i in nums1:        map[i] = map[i]+1 if i in map else 1    for j in nums2:        if j in map and map[j] > 0:            res.append(j)            map[j] = 0    return resSolution 4:sort the two list, and use two pointer to search in the lists to find common elements.class Solution(object):def intersection(self, nums1, nums2):    """    :type nums1: List[int]    :type nums2: List[int]    :rtype: List[int]    """    res = []    nums1.sort()    nums2.sort()    i = j = 0    while (i < len(nums1) and j < len(nums2)):        if nums1[i] > nums2[j]:            j += 1        elif nums1[i] < nums2[j]:            i += 1        else:            if not (len(res) and nums1[i] == res[len(res)-1]):                res.append(nums1[i])            i += 1            j += 1    return res