Leetcode question brushing Diary - 15 Sum of three

Give you one containing n Array of integers nums, Judge nums Are there three elements in a,b,c , bring a + b + c = 0 ？ Please find out all and for 0 Triple without repetition .

Be careful ：
Answer cannot contain duplicate triples .

Example 1：

Input ：nums = [-1,0,1,2,-1,-4]
Output ：[[-1,-1,2],[-1,0,1]]

Example 2：

Input ：nums = []
Output ：[]

Example 3：

Input ：nums = [0]
Output ：[]

Tips ：
0 <= nums.length <= 3000
-105 <= nums[i] <= 105

Their thinking ：

First step ： First, judge whether the length of the passed in array is greater than or equal to 3, Or judge whether the array is empty , If the array length is less than 3 Or the array is empty , Then directly return an empty list .
The second step ： Sort the array , Can be called directly Arrays Sorting provided by tool class API Interface sort（） Method .
The third step ： Go through every number , Then take three of them and add them , Judge whether the sum is zero , Zero use HashSet For storage （HashSet Class cannot store duplicate numbers , This can solve the problem of de duplication ）
Step four ： take HashSet convert to List, And return data .

class Solution {
    
    public List<List<Integer>> threeSum(int[] nums) {
    
        List<List<Integer>> lists = new ArrayList<>();
        if(nums.length < 3 || nums == null){
    
            return lists;
        }
        Arrays.sort(nums);
        HashSet<List<Integer>> set = new HashSet<>();
        for (int i = 0; i < nums.length; i++) {
    
            if(nums[i] > 0){
    
                break;
            }
            int L = i+1, R = nums.length - 1;
            while (L < R){
    
                int sum = nums[i] + nums[L] + nums[R];
                if (sum == 0){
    
                    set.add(Arrays.asList(nums[i], nums[L++], nums[R--]));
                }else if(sum < 0){
    
                    L ++;
                }else if(sum > 0){
    
                    R --;
                }
            }
        }
        for (List<Integer> list: set) {
    
            lists.add(list);
        }
        return lists;
    }
}

Their thinking 2：
The basic idea is consistent with the above , Just don't use HashSet Deduplication , Instead, use array subscripts for de duplication .

class Solution {
    
    public List<List<Integer>> threeSum(int[] nums) {
    
        int n = nums.length;
        Arrays.sort(nums);
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        //  enumeration  a
        for (int first = 0; first < n; ++first) {
    
            //  It needs to be different from the last enumeration 
            if (first > 0 && nums[first] == nums[first - 1]) {
    
                continue;
            }
            // c  The corresponding pointer initially points to the rightmost end of the array 
            int third = n - 1;
            int target = -nums[first];
            //  enumeration  b
            for (int second = first + 1; second < n; ++second) {
    
                //  It needs to be different from the last enumeration 
                if (second > first + 1 && nums[second] == nums[second - 1]) {
    
                    continue;
                }
                //  Need assurance  b  The pointer is in  c  On the left side of the pointer 
                while (second < third && nums[second] + nums[third] > target) {
    
                    --third;
                }
                //  If the pointers coincide , With  b  Subsequent additions 
                //  There will be no satisfaction  a+b+c=0  also  b<c  Of  c  了 , You can exit the loop 
                if (second == third) {
    
                    break;
                }
                if (nums[second] + nums[third] == target) {
    
                    List<Integer> list = new ArrayList<Integer>();
                    list.add(nums[first]);
                    list.add(nums[second]);
                    list.add(nums[third]);
                    ans.add(list);
                }
            }
        }
        return ans;
    }
}

Their thinking 3：

class Solution {
    
    public List<List<Integer>> threeSum(int[] nums) {
    
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        for(int k = 0; k < nums.length - 2; k++){
    
            if(nums[k] > 0) break;
            if(k > 0 && nums[k] == nums[k - 1]) continue;
            int i = k + 1, j = nums.length - 1;
            while(i < j){
    
                int sum = nums[k] + nums[i] + nums[j];
                if(sum < 0){
    
                    while(i < j && nums[i] == nums[++i]);
                } else if (sum > 0) {
    
                    while(i < j && nums[j] == nums[--j]);
                } else {
    
                    res.add(new ArrayList<Integer>(Arrays.asList(nums[k], nums[i], nums[j])));
                    while(i < j && nums[i] == nums[++i]);
                    while(i < j && nums[j] == nums[--j]);
                }
            }
        }
        return res;
    }
}