双数之和+三数之和+四数之和

1.题目分析

2.算法原理

我们只需要利用双指针算法就可以计算排序之后的双数之和。排序用库中自带的sort函数。双指针算法利用下标访问。用int和下标访问代替指针更为直观。还需要考虑去重的操作。即left++之后，numsleft==numsleft-1还需要left++;

3.代码实操

class Solution {
public:
    vector<int> twoSum(vector<int>& price, int target) {
        vector<int>nums;
        size_t left=0;
        size_t right=price.size()-1;;
        while(left<right)
        {
            if(price[left]+price[right]<target)
            {
                left++;
            }
            else if(price[left]+price[right]>target)
        {
            right--;
        }
        else
        {
               nums.push_back(price[left]);
               nums.push_back(price[right]);
                 return nums;
        }
        }
        return nums;
    }
};

1.题目分析

2.算法原理

三数之和先固定第一个数字，然后target-固定的数字取相反数等于两数之和即可，需要注意去重操作和双数之和类似。

3.代码实操

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
size_t left = 1;
int right = nums.size() - 1;
size_t cur = 0;
int count = 0;
vector<vector<int>>num;
vector<int>arr;
while (cur != nums.size() - 2)
{
	while (left < right)
	{
		if (nums[left] + nums[right] == -nums[cur])
		{
			arr.push_back(nums[left]);
			arr.push_back(nums[right]);
			arr.push_back(nums[cur]);
			num.push_back(arr);
			arr.clear();
			 count = 0;
			while (nums[left] == nums[left + 1])
			{
				if (left + 1 < nums.size() )
					left++;
				if (left + 1 == nums.size())
					break;
			}
			if (left + 1 < nums.size() )
				left++;
			else
				break;
			while (nums[right] == nums[right - 1])
			{
				if (right - 1 > -1)
				right--;
				if (right - 1 == -1)
					break;
			}
			if (right - 1 > -1)
				right--;
			else
				break;
		}
		else
		{
			if (nums[left] + nums[right] < -nums[cur])
			{
				if (left + 1 < nums.size() )
				left++;
			}
			else {
				if (right - 1 > -1)
				right--;
			}
		}
	}
	while (nums[cur] == nums[cur + 1])
	{
		if(cur+1<nums.size())
		cur++;
		if (cur == nums.size() - 2)
			break;
	}
	if (cur == nums.size() - 2)
		break;
	if (cur + 1 < nums.size())
		cur++;
	else
		break;
	right = nums.size() - 1;
	left = cur + 1;
}
return num;
    }
};

1.题目分析

2.算法原理

四数之和用三数之和和双数之和。依次固定第一个数字第二个数字，然后用双数之和。它们的操作都是类似的。区别在于这一个题目用前一个题目的思路。为了避免数字相同超出边界要做边界判断。

注意：去重操作和不漏操作。

去重在于++之后看是否和前一个值相等，是则再次++否则结束循环。

不漏操作时找出双数之和和三数之和后还要继续该循环，找出所以得双数之和和三数之和的符合值。

3.代码实操

class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
    	vector<int>a;
	vector<vector<int>>num;
        if(nums.size()<4) return num;
	sort(nums.begin(), nums.end());
	int cur = 0;
	int pos = 1;
	int left = pos + 1;
	int right = nums.size() - 1;
	int sum = 0;
	while (cur < nums.size() - 3)
	{
		long long int target1 = target - nums[cur];
		while (pos < nums.size() - 2)
		{
			long long int target2 = target1 - nums[pos];
			while (left < right)
			{
				sum = nums[left] + nums[right];
				if (sum < target2)
				{
					left++;
				}
				if (sum > target2)
				{
					right--;
				}
				if (sum == target2)
				{
					a.push_back(nums[cur]);
					a.push_back(nums[pos]);
					a.push_back(nums[left]);
					a.push_back(nums[right]);
					num.push_back(a);
					a.clear();
					left++;
					right--;
					while (left < right&&nums[left - 1] == nums[left])
					{
						left++;
					}
					while (left < right&&nums[right + 1] == nums[right])
					{
						right--;
					}
				}
			}
			pos++;
			while (pos<nums.size()-2&&nums[pos - 1] == nums[pos])
			{
				pos++;
				if (pos >= nums.size() - 2)break;
			}
			if (pos >= nums.size() - 2)break;
			left = pos + 1;
			right = nums.size() - 1;
		}
		cur++;
		while (nums[cur - 1] == nums[cur])
		{
			cur++;
			if (cur >= nums.size() - 3)break;
		}
		if (cur >= nums.size() - 3)return num;
		pos = cur + 1;
		left = pos + 1;
		right = nums.size() - 1;
	}
	return num;
}
};