LeetCode(18)-4S

版权声明：XiangYida https://blog.csdn.net/qq_36781505/article/details/83593590

4Sum

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]
]

嗯？这个题想了一种觉得比较简单的算法但是超时了

public List<List<Integer>> fourSum(int[] nums, int target) {
       Arrays.sort(nums);
       List<List<Integer>>lists=new LinkedList<>();
       boolean[]judge=new boolean[nums.length];
       int size=nums.length;
       for (int i = 0; i <size; i++) {
           judge[i]=true;
           for (int j = 0; j <size; j++) {
               if(judge[j])continue;
               judge[j]=true;
               for (int k = 0; k <size; k++) {
                   if(judge[k])continue;
                   judge[k]=true;
                   for (int l = 0; l <size; l++) {
                       if(judge[l])continue;
                       if(nums[i]+nums[j]+nums[k]+nums[l]==target){
                           List<Integer>list=new LinkedList<>();
                           list.add(nums[i]);
                           list.add(nums[j]);
                           list.add(nums[k]);
                           list.add(nums[l]);
                           Collections.sort(list);
                           if(!lists.contains(list))
                           lists.add(list);
                       }
                   }
                  judge[k]=false;
               }
               judge[j]=false;
           }
           judge[i]=false;
       }
       return lists;
   }

看了一下测试用例，估计是故意的,尴尬

但我觉得，四重循环没毛病，我加了些判断减少了大部分循环次数，主要是判断重复太费时间了，后面我再来看看没有机会优化一下吧，最后贴上大神的代码吧，用例14ms当真是优秀

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;
}

同样是四重循环，怎么差距就这么大呢？尴尬