4Sum(4个数的和为target的集合)leetcode18

骑猪看日落 2022-05-23 20:20 111阅读 0赞

Given an array `nums` of *n* integers and an integer `target`, are there elements *a*, *b*, *c*, and *d* in `nums` 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.

Example:

Given array nums = [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]
    ]

code1：（暴力一点的）

List<List<Integer>> result=new ArrayList<>();
        public List<List<Integer>> fourSum(int[] nums, int target) {
            if(nums.length<4)
                return result;
            Arrays.sort(nums);
            for(int i=0;i<=nums.length-4;){
                for(int j=i+1;j<=nums.length-3;){
                    int front=j+1;
                    int last=nums.length-1;
                    while(front<last){
                        int sum=nums[front]+nums[last]+nums[i]+nums[j];
                        if(sum==target){
                            List<Integer> list=new ArrayList<>();
                            list.add(nums[i]);
                            list.add(nums[j]);
                            list.add(nums[front]);
                            list.add(nums[last]);
                            result.add(list);
                            front++;
                            while(front<last&&nums[front]==nums[front-1])
                                front++;
                            last--;
                            while(front<last&&nums[last]==nums[last+1])
                                last--;
                        }else if(sum<target){
                            front++;
                            while(front<last&&nums[front]==nums[front-1])
                                front++;
                        }else{
                            last--;
                            while(front<last&&nums[last]==nums[last+1])
                                last--;
                        }
                    }
                    j++;
                    while(j<nums.length&&nums[j]==nums[j-1])
                        j++;
                }
                i++;
                while(i<nums.length&&nums[i]==nums[i-1])
                        i++;
            }
            return result;
        }

code2（优化后的）

public List<List<Integer>> fourSum(int[] nums, int target) {
    		ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
    		int len = nums.length;
    		if (nums == null || len < 4)
    			return res;
    
    		Arrays.sort(nums);
    
    		int max = nums[len - 1];
    		if (4 * nums[0] > target || 4 * max < target)
    			return res;
    
    		int i, z;
    		for (i = 0; i < len; i++) {
    			z = nums[i];
    			if (i > 0 && z == nums[i - 1])// avoid duplicate
    				continue;
    			if (z + 3 * max < target) // z is too small
    				continue;
    			if (4 * z > target) // z is too large
    				break;
    			if (4 * z == target) { // z is the boundary
    				if (i + 3 < len && nums[i + 3] == z)
    					res.add(Arrays.asList(z, z, z, z));
    				break;
    			}
    
    			threeSumForFourSum(nums, target - z, i + 1, len - 1, res, z);
    		}
    
    		return res;
    	}
    
    	/*
    	 * Find all possible distinguished three numbers adding up to the target
    	 * in sorted array nums[] between indices low and high. If there are,
    	 * add all of them into the ArrayList fourSumList, using
    	 * fourSumList.add(Arrays.asList(z1, the three numbers))
    	 */
    	public void threeSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,
    			int z1) {
    		if (low + 1 >= high)
    			return;
    
    		int max = nums[high];
    		if (3 * nums[low] > target || 3 * max < target)
    			return;
    
    		int i, z;
    		for (i = low; i < high - 1; i++) {
    			z = nums[i];
    			if (i > low && z == nums[i - 1]) // avoid duplicate
    				continue;
    			if (z + 2 * max < target) // z is too small
    				continue;
    
    			if (3 * z > target) // z is too large
    				break;
    
    			if (3 * z == target) { // z is the boundary
    				if (i + 1 < high && nums[i + 2] == z)
    					fourSumList.add(Arrays.asList(z1, z, z, z));
    				break;
    			}
    
    			twoSumForFourSum(nums, target - z, i + 1, high, fourSumList, z1, z);
    		}
    
    	}
    
    	/*
    	 * Find all possible distinguished two numbers adding up to the target
    	 * in sorted array nums[] between indices low and high. If there are,
    	 * add all of them into the ArrayList fourSumList, using
    	 * fourSumList.add(Arrays.asList(z1, z2, the two numbers))
    	 */
    	public void twoSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,
    			int z1, int z2) {
    
    		if (low >= high)
    			return;
    
    		if (2 * nums[low] > target || 2 * nums[high] < target)
    			return;
    
    		int i = low, j = high, sum, x;
    		while (i < j) {
    			sum = nums[i] + nums[j];
    			if (sum == target) {
    				fourSumList.add(Arrays.asList(z1, z2, nums[i], nums[j]));
    
    				x = nums[i];
    				while (++i < j && x == nums[i]) // avoid duplicate
    					;
    				x = nums[j];
    				while (i < --j && x == nums[j]) // avoid duplicate
    					;
    			}
    			if (sum < target)
                 while(i<j&&nums[++i]==nums[i-1])
                        ;
    			if (sum > target)
    				 while(i<j&&nums[--j]==nums[j+1])
                        ;
    		}
    		return;
    	}

code3：（优化后的代码短的）

public List<List<Integer>> fourSum(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(num[i]+num[i+1]+num[i+2]+num[i+3]>target)break; //first candidate too large, search finished
                if(num[i]+num[num.length-1]+num[num.length-2]+num[num.length-3]<target)continue; //first candidate too small
                if(i>0&&num[i]==num[i-1])continue; //prevents duplicate result in ans list
                for(int j=i+1; j<num.length-2; j++){
                    if(num[i]+num[j]+num[j+1]+num[j+2]>target)break; //second candidate too large
                    if(num[i]+num[j]+num[num.length-1]+num[num.length-2]<target)continue; //second candidate too small
                    if(j>i+1&&num[j]==num[j-1])continue; //prevents duplicate results in ans list
                    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++; //skipping over duplicate on low
                            while(low<high&&num[high]==num[high-1])high--; //skipping over duplicate on high
                            low++; 
                            high--;
                        }
                        //move window
                        else if(sum<target)
                            while(low<high&&num[++low]==num[low-1])
                                ;
                        else
                            while(low<high&&num[--high]==num[high+1])
                                ; 
                       
                    }
                }
            }
            return ans;
        }