406 double pointer (27. remove elements, 977. square of ordered array, 15. sum of three numbers, 18. sum of four numbers)

27. Remove elements

class Solution {
    
public:
    int removeElement(vector<int>& nums, int val) {
    
        
        int i = 0;
        int j = 0;

        for(; i < nums.size(); i++)
        {
    
            if(nums[i] == val)
            {
    
                continue;
            }

            nums[j++] = nums[i];
        }
        return j;
    }
};

977. The square of an ordered array

class Solution {
    
public:
    vector<int> sortedSquares(vector<int>& nums) {
    
        vector<int> v(nums.size(), 0);

        for(int i = 0; i < nums.size(); i++)
        {
    
            v[i] = nums[i]*nums[i];
        }

        sort(v.begin(), v.end());
        return v;
    }
};

15. Sum of three numbers

class Solution {
    
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
    

        vector<vector<int>> result;

        sort(nums.begin(), nums.end());


        for (int i = 0; i < nums.size(); i++)
        {
    
            if (nums[i] > 0)
            {
    
                return result;
            }

            if (i > 0 && nums[i] == nums[i - 1])
            {
    
                continue;
            }

            int left = i + 1;
            int right = nums.size() - 1;

            while (right > left)
            {
    
                if (nums[i] + nums[left] + nums[right] > 0)
                {
    
                    right--;
                    while (right > left && nums[right] == nums[right + 1]) right--;
                }
                else if (nums[i] + nums[left] + nums[right] < 0)
                {
    
                    left++;
                    while (right > left && nums[left] == nums[left - 1])   left++;
                }
                else
                {
    
                    result.push_back(vector<int>{
    nums[i], nums[left], nums[right]});
                    // duplicate removal , stay i Without change , Only right-- and left++ It is possible to get the next sum 0 Array of ,nums Is increasing ！
                    while (left < right && nums[right] == nums[right-1]) right--;
                    while (left < right && nums[left] == nums[left+1]) left++;

                    right--;
                    left++;
                }
            }
        }
        return result;
    }
};

18. Sum of four numbers

class Solution {
    
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
    
        vector<vector<int>> result;
        sort(nums.begin(), nums.end());

        //nums[k]
        for (int k = 0; k < nums.size(); k++)
        {
    
            // Pruning treatment 
            if (nums[k] > target && (nums[k] >= 0 || target >= 0))
            {
    
                break;
            }

            // duplicate removal 
            if (k > 0 && nums[k] == nums[k - 1])
            {
    
                continue;
            }

            //nums[i]
            for (int i = k + 1; i < nums.size(); i++)
            {
    
                // Secondary pruning treatment 
                if (nums[k] + nums[i] > target && (nums[k] + nums[i] >= 0 || target >= 0))
                {
    
                    break;
                }

                // The right way to get rid of duplication 
                if (i > k + 1 && nums[i] == nums[i - 1])
                {
    
                    continue;
                }

                int left = i + 1;
                int right = nums.size() - 1;

                while (right > left)
                {
    
                    if (nums[k] + nums[i] > target - (nums[left] + nums[right]))
                    {
    
                        right--;
                        while (right > left && nums[right] == nums[right + 1]) right--;
                    }
                    else if (nums[k] + nums[i] < target - (nums[left] + nums[right]))
                    {
    
                        left++;
                        while (right > left && nums[left] == nums[left - 1])   left++;
                    }
                    else
                    {
    
                        result.push_back(vector<int>{
    nums[k], nums[i], nums[left], nums[right]});

                        while (right > left && nums[left] == nums[left + 1]) left++;
                        while (right > left && nums[right] == nums[right - 1]) right--;

                        right--;
                        left++;
                    }

                }
            }
        }
        return result;
    }
};