*15. 3Sum (three pointers to two pointers), hashset

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]
]

Idea: Simplify this problem from O(n^3) to n square by using two pointers

iterate i and j = i+1 and k = n-1 (j and k are two pointers we would use)

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        //two pointers
        //set i and move j and k (if sum <0 j++ else k--)
        for(int i = 0; i<nums.length; i++){
            if (i != 0 && nums[i] == nums[i - 1]) continue; //*****
            int j = i+1;
            int k = nums.length-1;
            while(j<k){
                if(nums[i] + nums[j] + nums[k] == 0){
                    List<Integer> temp = new ArrayList<Integer>();
                    temp.add(nums[i]); temp.add(nums[j]);temp.add(nums[k]);
                    //if(!res.contains(temp))  //why add this make TLE ****
                        res.add(temp);
                    ++j;
                    //System.out.println("wei");
                    while (j < k && nums[j] == nums[j-1]) ++j; ****
                }else if(nums[i] + nums[j] + nums[k] < 0){
                    j++;
                }else {
                    k--;
                }
            }
        }
        return res;
    }
}

1.avoid the duplicate elements -1 -1 (for the same values, there are same results)

2. avoid using contains because of O(n), that is the reason why we need check the duplicate elements manually instead of using contains

Solution 2: using hashmap: n^2*lgn

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        for(int i = 0; i < nums.length; i++){
            if(i!=0 && nums[i-1] == nums[i]) continue;
            Set<Integer> set = new HashSet<>(); // no duplicate elements
            for(int j = i+1; j<nums.length; j++){// nums[j] : b and c means: count all nums[i] as c
                if(set.contains(-nums[i]-nums[j])){ // c
                    List<Integer> temp = new ArrayList<>();
                    temp.add(nums[i]);temp.add(nums[j]);temp.add(-nums[i]-nums[j]);
                    res.add(temp);
                    //avoid the duplicate elemnts
                    ++j;
                    while(j < nums.length && nums[j-1]==nums[j]) j++;
                    --j;
                }
                if(j<nums.length)
                set.add(nums[j]);
            }
        }
        return res;
    }
}

hashset

how to using two loop to represent three numbers.

1. treat all nums[j] as c(the third elemnts)

2. As a+b+c = 0, c = -a-b, we need find a and b to satisfy the requirement

原文地址：https://www.cnblogs.com/stiles/p/leetcode15.html