leetcode-18 4Sum



问题描述：Givenan array S of n integers, are there elements a, b, c,and d in S such that a + b + c + d =target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d)must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)

• The solution set must not containduplicate quadruplets.

    For example, given array S = {1 0 -1 0 -22}, and target = 0.

    A solution set is:

    (-1, 0, 0, 1)

    (-2, -1, 1, 2)

    (-2, 0, 0, 2)

问题分析：KSum问题，往下最终转化为基本的2Sum问题，同时注意消除重复值即可

代码：

java解法：

public class Solution {
    public List<List<Integer>> fourSum(int[] num, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
		if(num == null || num.length <= 3)
			return result;
		
		//先对数组进行排序
		Arrays.sort(num);
		
		int temp_target = 0;//每一次转化为2Sum问题的目标值
		for(int i = 0; i < num.length - 3; i++)
		{
			//消除重复值
			while((i != 0) && (i < num.length - 3) && (num[i] == num[i - 1]))
				++ i;
			//消除重复值之后一定要注意数组边界的判断，避免出现越界情况
			if(i < num.length - 3)
			{
				for(int j = i + 1; j < num.length - 2; j++)
				{
					while((j != i + 1) && (j < num.length - 2) && (num[j] == num[j - 1]))
						++ j;
					if(j < num.length - 2)
					{
						temp_target = target - num[i] - num[j];
						int start = j + 1;
						int end = num.length - 1;
				
						while(start < end)
						{
							int temp_sum = num[start] + num[end]; 
							if(temp_sum < temp_target)
							{
								++ start;
							}
							else if(temp_sum > temp_target)
							{
								-- end;
							}
							else
							{
								List<Integer> list = new ArrayList<>();
								list.add(num[i]);
								list.add(num[j]);
								list.add(num[start++]);
								list.add(num[end--]);
						
								result.add(list);
						
								//继续往前搜索，并消除相同值
								while((start < end) && (num[start] == num[start - 1]))
									++ start;
								while((start < end) && (num[end] == num[end + 1]))
									-- end;
							}
						}	
					}
				}
			}	
		}
		return result;	
    }
}

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Note: The Solution object is instantiated only once.
        vector<vector<int>> res;
    	int numlen = num.size();
		if(num.size()<4) return res;
		
		sort(num.begin(),num.end());
		set<vector<int>> tmpres;
		for(int i = 0; i < numlen; i++)
		{
			for(int j = i+1; j < numlen; j++)
			{
				int begin = j+1;
				int end = numlen-1;
				while(begin < end)
				{
					int sum = num[i]+ num[j] + num[begin] + num[end];
					if(sum == target)
					{
						vector<int> tmp;
						tmp.push_back(num[i]);
						tmp.push_back(num[j]);
						tmp.push_back(num[begin]);
						tmp.push_back(num[end]);
						tmpres.insert(tmp);
						begin++;
						end--;
					}else if(sum<target)
						begin++;
					else
						end--;
				}
			}
		}
		set<vector<int>>::iterator it = tmpres.begin();
		for(; it != tmpres.end(); it++)
			res.push_back(*it);
		return res;
    }
};