[Leetcode] 3Sum / 3Sum Closet

3SumJan 18 '128685 / 33750

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ? b ? c)
• The solution set must not contain duplicate triplets.

 

    For example, given array S = {-1 0 1 2 -1 -4},

    A solution set is:
    (-1, 0, 1)
    (-1, -1, 2)

 

 

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        sort(num.begin(), num.end());
        vector<vector<int> > res;
        int len = num.size();
        for (int i = 0; i < len - 2; i++) {
            if (i > 0 && num[i] == num[i-1]) continue;
            int a = i+1, b = len -1; 
            int t = -num[i];
            while (a < b) {
                if (num[a] + num[b] == t) {
                    int tmp[3] = {-t, num[a], num[b]};
                    vector<int> tmp2(tmp, tmp+3);
                    res.push_back(std::move(tmp2));
                    while (a < b && num[a] == num[a+1]) a++;
                    a++;
                    while (a < b && num[b] == num[b-1]) b--;
                    b--;
                }
                else if (num[a] + num[b] > t) {
                    b--;
                } else  a++;
            }
        }
        return res;
    }
};

 

 

3Sum ClosestJan 18 '125336 / 13935

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

    For example, given array S = {-1 2 1 -4}, and target = 1.

    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

» Solve this problem

 

 

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        sort(num.begin(), num.end());
        int len = num.size();
        int mindelta = 1 << 30;
        int res;
        for (int i = 0; i < len - 2; i++) {
            if (i > 0 && num[i] == num[i-1]) continue;
            int a = i+1, b = len -1; 
            int t = target - num[i];
            while (a < b) {
                int delta =  t - num[a] - num[b]; 
                if (delta == 0) return target;
                else if (delta > 0) {
                    if (delta < mindelta) mindelta = delta, res = target - delta;
                    a++;
                } else {
                    if (-delta < mindelta) mindelta = -delta, res = target- delta;
                    b--;
                }
            }
        }
        return res;
    }
};