LeetCode18--4Sum--数组中某四个元素之和为某个输入的数值，输出这四个元素的值，并且这个四元组唯一

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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]]

public class Solution {    public List<List<Integer>> fourSum(int[] nums, int target) {        Arrays.sort(nums);        List<List<Integer>> res = new LinkedList<>();//保证输出的顺序就是插入的顺序        if(nums == null || nums.length <= 0){            return res;        }        int max = nums[nums.length - 1];        for (int i = 0; i < nums.length - 3; i++) {//比如：{-4, -1, -1, 0, 1, 5 ,5},最后一次比较的是0,1,5,5            if (4 * nums[0] > target || 4 * max < target) {                break;            }            if (nums[0] + 3 * nums[1] > target || max + 3 * nums[nums.length - 2] < target) {                break;            }            if (4 * nums[i] > target) {                break;            }            if (nums[i] + 3 * max < target) {                continue;            }            if(nums[i]*4 == target){                if(i + 3 < nums.length  && nums[i+3] == nums[i]){                    res.add(Arrays.asList(nums[i],nums[i],nums[i],nums[i]));                }                continue;            }            List<Integer> list;            for (int j = i + 1; j < nums.length - 2; j++) {                if (j == i + 1 || (j > i + 1 && nums[j] != nums[j - 1])) {//比如-1只能做基准1次，后面的-1直接跳过                    int low = j + 1;                    int high = nums.length - 1;//最后一次num[i] = 0,num[j] = 1,num[lo] = 5,num[hi] = 5                    while (low < high) {                        if (nums[i] + nums[j] + nums[low] + nums[high] == target) {                            list = Arrays.asList(nums[i], nums[j], nums[low], nums[high]);                            if (!res.contains(list)) {//去除重复的结果                                res.add(list);                            }                            while (low < high && nums[low] == nums[low + 1])                                low++;//比如：{-4, -1, -1, 0, 1, 5 ,5}，-1只能计算一次，不然会有重复计算                            while (low < high && nums[high] == nums[high - 1])                                high--;//比如：{-4, -1, -1, 0, 1, 5, 5}，5只能计算一次，不然会有重复计算                            low++;                            high--;                        } else if (nums[i] + nums[j] + nums[low] + nums[high] < target) {                            low++;                        } else                            high--;                    }//while                }//if            }//for        }//for        return res;    }//fourSum}

282 / 282 test cases passed.

Status: Accepted
Runtime: 36 ms
Your runtime beats 76.98 % of java submissions.
T(n) = O(n^3)

方法二：

public 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，num[i]不能和num[j]去比较是否相等                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;    }}

282 / 282 test cases passed.
Status: Accepted
Runtime: 24 ms

Your runtime beats 97.98 % of java submissions.

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