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.
Not good enough approach
var intersect = function(nums1, nums2) { let r = []; // get larger array const len1 = nums1.length; const len2 = nums2.length; // larger & smaller const larger = len1 > len2 ? nums1: nums2; const smaller = len1 > len2 ? nums2: nums1; // conver larger array to object let hashed = {}; for (let n of larger) { if (n in hashed) { hashed[n]++; } else { hashed[n] = 1; } } // loop over smaller array for (let n of smaller) { if (`${n}` in hashed) { r.push(n); hashed[n] = hashed[n]-1; if (hashed[n] === 0) { delete hashed[n]; } } } return r; };
The reason that code above is not good enough is because, \
1. we use larger array as lookup, this cause more memory usage. -- actually we need to use smaller array as lookup
2. we use ‘len1, len2, samller, larger‘ extra storage, we can actully swap nums1 and nums by one extra function call.
var intersect = function(nums1, nums2) { if (nums1.length > nums2.length) { return intersect(nums2, nums1); } // conver samller array to object let hashed = {}; for (let n of nums2) { if (n in hashed) { hashed[n]++; } else { hashed[n] = 1; } } let r = []; // loop over smaller array for (let n of nums1) { if (hashed[n] > 0) { r.push(n); hashed[n] = hashed[n]-1; } } return r; }
Follow up:
Will write another post for the follow up questions.
[Algorithm] 350. Intersection of Two Arrays II
原文：https://www.cnblogs.com/Answer1215/p/12040497.html