350. Intersection of Two Arrays II
# Easy
Two methods: (assume len(nums1) < len(nums2))
暴力解法. Ask each element in nums1 whether in nums2, if yes, then remove this element from both lists and add to result list.
Hash map. Build a dictinonary <numss[i], n>, means nums1[i] appears n times in nums1. Use this dictionary to traversal all elements in nums2.
Time complexity = if if x in nums2
only consume ; space complexity =
Last updated