[Leetcode] 4Sum

一、使用2sum降低复杂度为O（N^2）

public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        List<List<Integer>> res = new LinkedList<List<Integer>>();
        for(int i=0;i<nums.length-3;i++){
            if(i!=0&&nums[i-1]==nums[i]) continue;
            for(int j=i+1;j<nums.length-2;j++){
                if(j!=i+1&&nums[j-1]==nums[j]) continue;
                int sum = target - (nums[i]+nums[j]);
                int left = j+1;
                int right = nums.length-1;
                while(left<right){
                    int tmpsum = nums[left] + nums[right];
                    if(tmpsum==sum){
                        List<Integer>listone = new LinkedList<Integer>();
                        listone.add(nums[i]);
                        listone.add(nums[j]);
                        listone.add(nums[left]);
                        listone.add(nums[right]);
                        res.add(listone);
                        left++;
                        right--;
                        while(left<right&&nums[left-1]==nums[left]) left++;
                        while(left<right&&nums[right]==nums[right+1]) right--;
                    }else if(tmpsum < sum){
                        left++;
                        while(left<right&&nums[left-1]==nums[left]) left++;
                    }
                    else{
                        right--;
                        while(left<right&&nums[right]==nums[right+1]) right--;
                    }
                }
            }
        }
        return res;
    }
}

 

二、和3Sum是一样的，只是多了一层循环而已。

public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        List<List<Integer> > listall = new LinkedList<List<Integer> >();
        for(int i=0;i<nums.length;i++){
            int num1=nums[i];
            if(i>0&&num1==nums[i-1]) continue;
            for(int j=i+1;j<nums.length;j++){
                int num2=nums[j];
                if(j>i+1&&num2==nums[j-1]) continue;
                for(int k=j+1;k<nums.length;k++){
                    int num3=nums[k];
                    if(k>j+1&&num3==nums[k-1]) continue;
                    for(int m=k+1;m<nums.length;m++){
                        int num4=nums[m];
                        if(m>k+1&&num4==nums[m-1]) continue;
                        int sum=num1+num2+num3+num4;
                        if(sum==target){
                            List<Integer> listone = new LinkedList<Integer>();
                            listone.add(num1);
                            listone.add(num2);
                            listone.add(num3);
                            listone.add(num4);
                            listall.add(listone);
                        }
                        if(sum>target)
                            break;
                    }
                }
            }
        }
        return listall;
    }
}