Leetcode-015-三数之和

先排序，再用双指针，注意的是解中是没有重复的情况的，所以需要跳过。

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);
        
        for(int i=0;i<nums.length;i++){
            if(i>0&&nums[i]==nums[i-1]) continue;
            int l =i+1;
            int r= nums.length-1;
            
            int twoSum = 0-nums[i];
            
            while(l<r){
                
                if(nums[l]+nums[r]==twoSum){
                    res.add(Arrays.asList(nums[i], nums[l], nums[r]));
                    while(l<r&&nums[l+1]==nums[l]) l++;
                    while(l<r&&nums[r-1]==nums[r]) r--;
                    l++;
                    r--;
                }else if(nums[l]+nums[r]<twoSum) l++;
                else if(nums[l]+nums[r]>twoSum) r--;
            }
        }
        return res;                    
    }
}

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        res = []

        for i in range(len(nums)):
            if i>0 and nums[i]==nums[i-1]:
                continue;
            
            two_sum = - nums[i]
            l = i+1
            r = len(nums)-1
            while(l<r):
                if nums[l]+nums[r]==two_sum:
                    res.append([nums[i], nums[l], nums[r]])
                    while(l<r and nums[l]==nums[l+1]):l+=1
                    while(l<r and nums[r]==nums[r-1]):r-=1
                    l+=1
                    r-=1
                elif nums[l]+nums[r]<two_sum:l+=1
                else:r-=1
        return res