Leetcode 18 4Sum

最新推荐文章于 2021-12-16 21:15:08 发布

原创最新推荐文章于 2021-12-16 21:15:08 发布 · 236 阅读

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本文深入探讨了四数之和算法，通过两种不同的方法实现寻找数组中四个元素的组合，使其和等于目标值，同时避免重复组合。第一种方法采用双指针策略，结合HashSet去重；第二种方法优化搜索过程，避免重复计算，达到更高效的解决方案。

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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]
]

这个题跟前面的3Sum，3Sum closest很相似，这次采用了set去重，可以直接遍历，也可以设置双指针。

1）

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> list = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        int len = nums.length;
        Set<List> set = new HashSet<>();
        for(int i = 0 ; i < len - 3 ; i++){
            for(int j = i + 1 ; j < len - 2 ;j++){
                int left = j + 1;
                int right = len - 1;
                while(left < right){
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if(sum == target){
                        List<Integer> li = new ArrayList<>();
                        li.add(nums[i]);
                        li.add(nums[j]);
                        li.add(nums[left]);
                        li.add(nums[right]);
                        if(!set.contains(li)){
                            set.add(li);
                            list.add(li);
                        }
                        --right;
                        ++left;
                    }else if(sum > target){
                        right--;
                    }else{
                        left++;
                    }
                }
            }
        }
        return list;
    }
}

51ms

2）

class Solution {
    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)
                        low++; 
                    else 
                        high--;
                }
            }
        }
        return ans;
    }
}
//14毫秒的最佳结果。