Three different numbers with 0 in leetcode/ array

Code

package com.xcrj;

import java.util.*;

/** *  The finger of the sword  Offer II 007.  Array and  0  Three numbers of  *  Arrays are unordered  *  Given an inclusion  n  Array of integers nums, Judge nums Are there three elements in a ,b ,c , bring a + b + c = 0 ？ Please find all and for  0  Triple without repetition  * a + b + c = 0 And ask for a b c Are different  */
public class Solution7 {
    

    /** *  Convert to sword finger  Offer II 006.  Sort the sum of two numbers in the array   problem  *  Sort the array  * a + b = c */
    public List<List<Integer>> threeSum1(int[] nums) {
    
        //  result ,
        Set<List<Integer>> set = new HashSet<>(3);
        //  Ascending sort   above set？, The title requires that the results are not repeated , for example  -4,-1,-1,0,1,2.[-1,0,1],[-1,0,-1] It's repetitive 
        Arrays.sort(nums);
        //  Double pointer 
        for (int k = 0; k < nums.length; k++) {
    
            // target=-c
            int target = 0 - nums[k];
            //  from k The next element of starts applying double pointers 
            int i = k + 1;
            int j = nums.length - 1;
            while (i < j) {
    
                int sum = nums[i] + nums[j];
                if (sum == target) {
    
                    // asList(T... a)  Pass in variable parameters , Variable length parameter 
                    // !! add If the value is the same, discard  (e==null ? e2==null : e.equals(e2))
                    // !! ArrayList Of equal()  according to   Object references are equal || The element values in the array are equal 
                    set.add(Arrays.asList(nums[k], nums[i], nums[j]));
                    //  There are new a+b+c=0 The situation of 
                    i++;
                    j--;
                } else if (sum < target) i++;
                else j--;
            }
        }

        // ArrayList(Collection<? extends E> c)
        return new ArrayList<>(set);
    }

    /** *  Convert to sword finger  Offer II 006.  Sort the sum of two numbers in the array   problem  *  Sort the array  * a + b = c */
    public List<List<Integer>> threeSum2(int[] nums) {
    
        //  result ,
        List<List<Integer>> lists = new ArrayList<>(3);
        //  Ascending sort   above set？, The title requires that the results are not repeated , for example  -4,-1,-1,0,1,2.[-1,0,1],[-1,0,-1] It's repetitive 
        Arrays.sort(nums);
        //  Double pointer 
        for (int k = 0; k < nums.length; k++) {
    
            //  duplicate removal k,k+1 and k The values of the elements are the same , Handle 1 Just one , To deal with the first k Just one 
            if (k > 0 && nums[k] == nums[k - 1]) continue;

            // target=-c
            int target = 0 - nums[k];
            //  from k The next element of starts applying double pointers 
            int i = k + 1;
            int j = nums.length - 1;
            while (i < j) {
    
                int sum = nums[i] + nums[j];
                if (sum == target) {
    
                    // asList(T... a)  Pass in variable parameters , Variable length parameter 
                    lists.add(Arrays.asList(nums[k], nums[i], nums[j]));
                    //  duplicate removal , requirement a+b+c=0, a b c Are different 
                    while (i < j && nums[i] == nums[++i]) ;
                    while (i < j && nums[j] == nums[--j]) ;
                } else if (sum < target) i++;
                else j--;
            }
        }

        return lists;
    }

    public static void main(String[] args) {
    
        Solution7 solution7 = new Solution7();
        System.out.println(solution7.threeSum2(new int[]{
    -1, 0, 1, 2, -1, -4}));
    }
}

Reference resources

author ：tangweiqun
link ：https://leetcode.cn/problems/1fGaJU/solution/jian-dan-yi-dong-javac-pythonjs-san-shu-nu6el/
source ： Power button （LeetCode）