Leetcode-018-四数之和

Posted 2021-03-06 huangzengrui

和三数之和思路一样，先排序，再双指针。

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        if(nums.length==0){return res;}

        for(int i1=0;i1<nums.length-3;i1++){
            if(i1>0&&nums[i1-1]==nums[i1]){continue;}
            for(int i2=i1+1;i2<nums.length-2;i2++){
                if( i2>i1+1&&nums[i2-1]==nums[i2]){continue;}
                int i3=i2+1;
                int i4=nums.length-1;
                while(i3<i4){
                    if(nums[i1]+nums[i2]+nums[i3]+nums[i4]==target){
                        res.add(Arrays.asList(nums[i1],nums[i2],nums[i3],nums[i4]));
                        while(i3<i4&&nums[i3+1]==nums[i3]){i3++;}
                        while(i3<i4&&nums[i4-1]==nums[i4]){i4--;}
                        i3++;
                        i4--;
                    }else if(nums[i1]+nums[i2]+nums[i3]+nums[i4]<target){
                        i3++;
                    }else{
                        i4--;
                    }
                }
            }
        }
        return res;
    }
}

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        if not nums:return []

        nums.sort()

        res = []
        for i1 in range(len(nums)-3):
            if i1>0 and nums[i1]==nums[i1-1]:
                continue
            for i2 in range(i1+1, len(nums)-2):
                if i2>i1+1 and nums[i2]==nums[i2-1]:
                    continue
                i3 = i2+1
                i4 = len(nums)-1
                while i3<i4:
                    if nums[i1]+nums[i2]+nums[i3]+nums[i4]==target:
                        res.append([nums[i1],nums[i2],nums[i3],nums[i4]])
                        while i3<i4 and nums[i3+1]==nums[i3]:
                            i3+=1
                        while i3<i4 and nums[i4-1]==nums[i4]:
                            i4-=1
                        i3+=1
                        i4-=1
                    elif nums[i1]+nums[i2]+nums[i3]+nums[i4]<target:
                        i3+=1
                    else:
                        i4-=1
        return res