15 3sum

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]

对比2Sum，可以将3Sum分解为遍历a加上b+c=-a来考虑。
先对数组进行排序，然后用两指针向中心逼近来求2Sum。
此题可借以练习排序。

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        
        int length=nums.length;
        int i=0;
        int lo=0,hi=0,sum=0;
        List<List<Integer>> res=new ArrayList<>();
        for(;i<length-2;i++){
            if(i==0||(nums[i]!=nums[i-1])){
                lo=i+1;
                hi=length-1;
                sum=0-nums[i];
                while(lo<hi){
                    if(nums[lo]+nums[hi]==sum){
                        res.add(Arrays.asList(nums[i],nums[lo],nums[hi]));
                        while(lo<hi&&nums[lo]==nums[lo+1]) lo++;
                        while(lo<hi&&nums[hi]==nums[hi-1]) hi--;
                        lo++;
                        hi--;
                    }else if(nums[lo]+nums[hi]<sum){
                        lo++;
                    }else{
                        hi--;
                    }
                }
            }
        }
        return res;
    }
    private void sort(int[]nums){
        //可自己选择实现排序
    }
}