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解題 8. Intersection of Two Arrays II

21 12 月, 2021 欣迪

出處： https://leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/674/

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.

Ex1.
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]

Ex2.
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 1000

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?

var intersect = function(nums1, nums2) {
    const shortArray = nums1.length < nums2.length ? nums1 : nums2
    const longArray = nums1.length < nums2.length ? nums2 : nums1
    return shortArray.filter((item, idx) => {
        if(longArray.indexOf(item) >= 0) {
            longArray[longArray.indexOf(item)] = null
            return true
        } else {
            return false
        }
    })
};

技巧

這題有幾個點會特別容易忽略

• 已經被配對過的文字不能再重複使用，可以修改陣列的內容為 null 避免重複配對，[1], [1,1] 回傳的結果應該要是 [1]
• 兩個陣列的長短攸關解題的結果和效能，使用短陣列做 filter 可以減少進入 loop 的次數

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