4 Sum

最新推荐文章于 2019-03-25 21:49:02 发布

Daisy_HJL

最新推荐文章于 2019-03-25 21:49:02 发布

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分类专栏： Algorithm 文章标签： four Sum

本文链接：https://blog.youkuaiyun.com/qq_28618765/article/details/72519190

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本文探讨了如何寻找数组中四个元素之和等于特定目标值的所有唯一组合，并提供了两种实现方法。第一种方法使用三个循环进行遍历，第二种方法则采用两数之和逐步扩展到四数之和的方式，避免重复并确保解集不包含重复的四元组。

Given an 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: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

有两种方法：一种是通过三个循环求解；另一种是用两位数的和，推到三位数的和，再推至四位数的和。

代码一：

public List<List<Integer>> fourSum2(int[] num,int target){
		ArrayList<List<Integer>> ans = new ArrayList<>();
		if(num.length<4)
			return ans;
		Arrays.sort(num);
		
		for(int i=0;i<num.length-3;i++){
			if(i>0 && num[i]==num[i-1])
				continue;
			for(int j=i+1;j<num.length-2;j++){
				if(j>i+1 && num[j]==num[j-1])
					continue;
				int low=j+1,high=num.length-1;
				while(low<high){
					int sum = num[i]+num[j]+num[low]+num[high];
					if(sum==target){
						ans.add(Arrays.asList(num[i],num[j],num[low],num[high]));
						while(low<high && num[low]==num[low+1])
							low++;
						while(low<high && num[high]==num[high-1])
							high--;
						low++;
						high--;
					}else if(sum<target){
						low++;
					}else{
						high--;
					}
				}
			}
		}
		return ans;		
	}

代码二：

public List<List<Integer>> fourSum(int[] nums, int target) {
        ArrayList<List<Integer>> ans = new ArrayList<>();
        if(nums == null || nums.length<4)
            return ans;
        Arrays.sort(nums);
        
        int max = nums[nums.length-1];
        if(4*nums[0] > target || 4*max<target)
            return ans;
        for(int i=0;i<nums.length-3;i++){
            int z = nums[i];
            if(i>0 && nums[i]==nums[i-1])
                continue;
            if(z+3*max <target)
                continue;
            if(4*max<target)
                break;
            if(4*max==target){
                if(i+3<nums.length && nums[i+3]==z){
                    ans.add(Arrays.asList(z,z,z,z));
                }
                break;
            }
            threeSumForFourSum(nums,target-z,i+1,nums.length-1,ans,z);
        }
        return ans;
    }
    
    public void threeSumForFourSum(int[] nums,int target,int low,int high,ArrayList<List<Integer>> ans,int z1){
        if(low+1>=high)
            return;
        int max = nums[high];
        
        if(3*nums[low]>target || 3*max<target)
            return;
        
        for(int i=low;i<high-1;i++){
            int z=nums[i];
            if(i>low && nums[i]==nums[i-1]){
                continue;
            }
            if(z+2*max<target){
                continue;
            }
            if(3*z>target){
                break;
            }
            if(3*z==target){
                if(i+1<high && nums[i+2]==z){
                    ans.add(Arrays.asList(z1,z,z,z));
                }
                break;
            }
                    
            twoSumForFourSum(nums,target-z,i+1,high,ans,z1,z);
        }
    }
    
    public void twoSumForFourSum(int[] nums,int target,int low,int high,ArrayList<List<Integer>> ans,int z1,int z2){
        if(low>=high)
            return;
        if(2*nums[low]>target || 2*nums[high] < target){
            return;
        }
        
        while(low<high){
            int sum=nums[low]+nums[high];
            if(sum == target){
                ans.add(Arrays.asList(z1,z2,low,high));
                
                while(low<high && nums[low]==nums[low+1])
                    low++;
                while(low<high && nums[high]==nums[high-1])
                    high--;
            }else if(sum > target){
                high--;
            }else{
                low++;
            }
        }
        return;
    }