Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
The solution set must not contain duplicate triplets.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
AC代码
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> ans;
if (nums.size() < 3) return ans;
set<vector<int>> st;
sort(nums.begin(), nums.end());
if (nums[0] == 0 & nums[nums.size() - 1] == 0) {
vector<int> tmp{0, 0, 0};
ans.push_back(tmp);
return ans;
}
for (auto mid = 1; mid < nums.size() - 1; mid++) {
int left = 0, right = nums.size() - 1;
while (left < mid && mid < right) {
if (nums[left] + nums[mid] + nums[right] == 0) {
vector<int> tmp{nums[left], nums[mid], nums[right]};
st.insert(tmp);
left++;
}
else if (nums[left] + nums[mid] + nums[right] > 0)
right--;
else if (nums[left] + nums[mid] + nums[right] < 0)
left++;
}
}
for (auto it = st.begin(); it != st.end(); ++it) {
vector<int> tmp{(*it)[0], (*it)[1], (*it)[2]};
ans.push_back(tmp);
}
return ans;
}
};
优化代码
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> res;
for (int i = 0; i < nums.size(); i++) {
if ((i > 0) && (i < nums.size()) && (nums[i] == nums[i - 1]))
continue;
int l = i + 1, r = nums.size() - 1;
while (l < r) {
int s = nums[i] + nums[l] + nums[r];
if (s > 0)
r--;
else if (s < 0)
l++;
else {
res.push_back(vector<int>{nums[i], nums[l], nums[r]});
while (l < r && nums[l] == nums[l + 1]) l++;
while (l < r && nums[r] == nums[r - 1]) r--;
l++;
r--;
}
}
}
return res;
}
};
总结
都是O(n2)的复杂度,自己写的效率还是低了,参考资料:https://www.acwing.com/solution/leetcode/content/1079/
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