3Sum

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)

---

分析：先对数组进行排序，先固定第一个数，然后两边夹逼。特别要注意的地方是 如何排除相同的结果。

进行排序后，连续的几个数字可能相同。因此，要略过相同的数字进行检查。
Java: run time 28ms, memory usage 44.3MB 

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer> > result = new ArrayList<>();
        int n = nums.length;
        int a = 0;
        Arrays.sort(nums);
        while (a < n - 2 && nums[a] <= 0) { // eliminate unnecessary checks
            if(a != 0 && nums[a] == nums[a - 1]) {
                a++;
                continue;
            }
            int b = a + 1;
            int c = n - 1;
            while(b < c) {
                int currentSum = nums[a] + nums[b] + nums[c];
                if(currentSum == 0) {
                    if(b != a + 1 && nums[b] == nums[b - 1]) {
                        b++;
                    } else if (c != n - 1 && nums[c] == nums[c + 1]) {
                        c--;
                    } else {
                        result.add(Arrays.asList(new Integer[] {nums[a], nums[b], nums[c]}));
                        b++;
                        c--;
                    }
                } else if (currentSum > 0) {
                    c--;
                } else {
                    b++;
                }
            }
            a++;
        }
        
        return result;
    }
}

 

C++: 运行时间65ms

class Solution {
public:
    vector<vector<int> > threeSum(vector<int>& nums) {
        vector<vector<int> > result;
        if(nums.size() < 3) return result;
        sort(nums.begin(), nums.end());
        
        for(int a = 0; a < nums.size() - 2; a++){
            if(a != 0 && nums[a] == nums[a-1]) continue;
            int b = a + 1, c = nums.size() - 1;
            while(b < c){
                if(nums[a] + nums[b] + nums[c] > 0) c--;
                else if(nums[a] + nums[b] + nums[c] < 0) b++;
                else{
                    if(b != a + 1 && nums[b] == nums[b-1]) b++;
                    else if(c != nums.size() - 1 && nums[c] == nums[c+1]) c--;
                    else{
                        result.push_back({nums[a], nums[b], nums[c]});
                        b++;
                        c--;
                    }
                }
            }
        }
        sort(result.begin(), result.end());
        return result;
    }
};