【组合&元素和】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)

DFS递归太深会超时，选用两层for循环

将3个和问题，转化为两个和的问题（两个指针指向数组）

注意去重：

• if (i > 0 或类似condition && num[i] == num[i - 1]) continue;

public class Solution {
    ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
    
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        if(num == null) return null;
        Arrays.sort(num);
        int len = num.length;
        
        for(int i=0; i<len; i++){
            if(i>0 && num[i] == num[i-1]) continue;//for No duplicate
            
            int p=i+1, q=len-1;
            while(p < q){//2 wei
                
                if(p!=i+1 && num[p]==num[p-1]){//for No duplicate
                    p++;
                    continue;
                }
                if(q!=len-1 && num[q]==num[q+1]){//for No duplicate
                    q--;
                    continue;
                }
                
                int sum = num[p] + num[q] + num[i];
                if(sum == 0){
                    ArrayList<Integer> l = new ArrayList<Integer>();
                    l.add(num[i]);
                    l.add(num[p]);
                    l.add(num[q]);
                    res.add(l);
                    p++;
                }
                else if(sum < 0) p++;
                else q--;
            }
        }
        
        return res;
    }
}