【LeetCode】排序

[349] Intersection of Two Arrays [Easy]

两个无序可重复数组找交集，交集要求元素唯一。

Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

思路：1、两个unordered_set 可以去重； 2、先排序，双指针

 1 //Space:O(N); Time: 0(N)
 2 //unordered_set insert, emplace, find, delete O(1)
 3 
 4 class Solution {
 5 public:
 6     vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
 7         unordered_set<int> st1;
 8         for (auto ele: nums1) {
 9             st1.emplace(ele);
10         }
11         unordered_set<int> st2;
12         for (auto ele: nums2) {
13             if (st1.find(ele) != st1.end()) {
14                 st2.emplace(ele);
15             }
16         }
17         vector<int> ans;
18         for (auto ele: st2) {
19             ans.push_back(ele);
20         }
21         return ans;
22     }
23 };

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 1 //space: no extra sapce, time(nlog(n))
 2 //two pointers
 3 class Solution {
 4 public:
 5     vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
 6         sort(nums1.begin(), nums1.end());
 7         sort(nums2.begin(), nums2.end());
 8         vector<int>::size_type idx1 = 0, idx2 = 0;
 9         vector<int> ans;
10         while(idx1 < nums1.size() && idx2 < nums2.size()) {
11             if(nums1[idx1] == nums2[idx2]) {
12                 if (ans.empty() || nums1[idx1] != ans.back()) { //使用api之前一定先做条件检测
13                     ans.push_back(nums1[idx1]);
14                 }
15                 idx1++, idx2++;
16             } else if (nums1[idx1] < nums2[idx2]) {
17                 ++idx1;
18             } else {
19                 ++idx2;
20             }
21         }
22         return ans;
23     }
24 };

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