15. 3Sum/三数之和

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> threeSum(vector& nums) {
        vector> ans;
        if (nums.size() < 3) return ans;
        set> st;
        sort(nums.begin(), nums.end());
        if (nums[0] == 0 & nums[nums.size() - 1] == 0) {
            vector 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 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 tmp{(*it)[0], (*it)[1], (*it)[2]};
            ans.push_back(tmp);
        }
        return ans;
    }
};

优化代码

class Solution {
public:
    vector> threeSum(vector& nums) {
        sort(nums.begin(), nums.end());
        vector> 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{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/