Code & Fun

第82天。

今天的题目是Intersection of Two Arrays:

Given two arrays, write a function to compute their intersection.

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

Note:
Each element in the result must be unique.
The result can be in any order.

可以用排序做,也可以用hash做:

排序的做法:

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vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
sort(nums1.begin(),nums1.end());
sort(nums2.begin(),nums2.end());
auto beg1 = nums1.begin();
auto beg2 = nums2.begin();
vector<int> ret;
while(beg1 < nums1.end() && beg2 < nums2.end()) {
if (*beg1 == *beg2) {
int t = *beg1;
ret.push_back(t);
while(beg1 < nums1.end() && *beg1 == t) beg1++;
while(beg2 < nums2.end() && *beg2 == t) beg2++;
} else if (*beg1 < *beg2) beg1++;
else beg2++;
}
return ret;
}

hash的做法:

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vector<int> intersection1(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int,int> m;
vector<int> ret;
for(auto i:nums1) m[i]++;
for(auto i:nums2)
if (m.find(i) != m.end() && m[i]) {
m[i] = 0;
ret.push_back(i);
}
return ret;
}

dicuss还有用set做的:

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vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
set<int> s(nums1.begin(), nums1.end());
vector<int> out;
for (int x : nums2)
if (s.erase(x))
out.push_back(x);
return out;
}

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