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)

 Idea:  o(n^2) solution exists. First sort the array, and then from left to right, for each num[i], search the pair that sums up to -num[i] using Two Sum algorithm. 

public class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        Arrays.sort(num);
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(num == null || num.length < 3) return result;
        int len = num.length;
        for(int i = 0; i < num.length-2; i ++){
            if(i==0 || num[i]>num[i-1]){
                int j = i + 1;
                int h = len - 1;
                while(j < h){
                    int sum = 0 - num[j] - num[h];
                    if(sum == num[i]){
                        ArrayList<Integer> row = new ArrayList<Integer>();
                        row.add(num[i]);
                        row.add(num[j]);
                        row.add(num[h]);
                        result.add(row);
                        h--;
                        j++;
                        while(h>j && num[h]==num[h+1]) h--; 

                        while(j<h && num[j]==num[j-1]) j++;
                    }else if(sum < num[i]){
                            h --;
                    }else if(sum > num[i]){
                            j ++;
                    }
                }
            }
        }
        return result;
    }
}

 第二遍：

public class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(num == null || num.length < 3) return result;
        Arrays.sort(num);
        for(int i = 0; i < num.length - 2; i ++){
            if(i == 0 || num[i] > num[i-1]){
                int target = - num[i];
                int j = i + 1;
                int h = num.length - 1;
                while(j < h){
                    if(num[j] + num[h] == target){
                        ArrayList<Integer> row = new ArrayList<Integer>();
                        row.add(num[i]);
                        row.add(num[j]);
                        row.add(num[h]);
                        result.add(row);
                        int jnum = num[j];
                        int hnum = num[h];
                        while(num[j] == jnum && j < h) j ++;
                        while(num[h] == hnum && h > j) h --;
                    }else if(num[j] + num[h] > target){
                        h --;
                    }else{
                        j ++;
                    }
                }
            }
        }
        return result;
    }
}