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| // @Title: 两个数组的交集 (Intersection of Two Arrays)
// @Author: 15816537946@163.com
// @Date: 2020-11-09 15:55:57
// @Runtime: 0 ms
// @Memory: 2.8 MB
/*
* @lc app=leetcode.cn id=349 lang=golang
*
* [349] 两个数组的交集
*
* https://leetcode-cn.com/problems/intersection-of-two-arrays/description/
*
* algorithms
* Easy (71.01%)
* Likes: 290
* Dislikes: 0
* Total Accepted: 139K
* Total Submissions: 189.9K
* Testcase Example: '[1,2,2,1]\n[2,2]'
*
* 给定两个数组,编写一个函数来计算它们的交集。
*
*
*
* 示例 1:
*
* 输入:nums1 = [1,2,2,1], nums2 = [2,2]
* 输出:[2]
*
*
* 示例 2:
*
* 输入:nums1 = [4,9,5], nums2 = [9,4,9,8,4]
* 输出:[9,4]
*
*
*
* 说明:
*
*
* 输出结果中的每个元素一定是唯一的。
* 我们可以不考虑输出结果的顺序。
*
*
*/
// @lc code=start
/*
func intersection(nums1 []int, nums2 []int) []int {
var intersection []int
set1 := map[int]struct{}{}
for _, v := range nums1 {
set1[v] = struct{}{}
}
set2 := map[int]struct{}{}
for _, v := range nums2 {
set2[v] = struct{}{}
}
if len(set1) > len(set2) {
set1, set2 = set2, set1
}
for v := range set1 {
if _, has := set2[v]; has {
intersection = append(intersection, v)
}
}
return intersection
}
*/
func intersection(nums1 []int, nums2 []int) (res []int) {
sort.Ints(nums1)
sort.Ints(nums2)
for i, j := 0, 0; i < len(nums1) && j < len(nums2); {
x, y := nums1[i], nums2[j]
if x == y {
if res == nil || x > res[len(res)-1] {
res = append(res, x)
}
i++
j++
} else if x < y {
i++
} else {
j++
}
}
return
}
|