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]
]

时间 o(n^2), space o(n)

不要忘记 先排序，然后循环的时候去重

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        
        if(nums == null || nums.length == 0){
            return res;
        }
        
        
        Arrays.sort(nums);
        for(int i = 0; i < nums.length - 2; i++){
            
            //去重
            if(i > 0 && nums[i] == nums[i-1]) continue;
            
            int low = i+1, hight = nums.length -1, sum = 0-nums[i];
            
            while(low < hight){
                if(nums[low] + nums[hight] == sum){
                    List<Integer> list = new ArrayList<>();
                    list.add(nums[low]);
                    list.add(nums[hight]);
                    list.add(nums[i]);
                    res.add(list);
                    //res.add(Array.asList())
                    //退出循环时，low在最后一个重复的数组，hight也是一样，所以最后要
                    while(low < hight && nums[low] == nums[low+1] ) low++;
                    while(low < hight && nums[hight] == nums[hight-1]) hight--;
                    
                    low++;
                    hight--;
                    
                }
                
                else if(nums[low] + nums[hight] < sum){
                    low++;
                }else{
                    hight--;
                }
            }
        }
        
        return res;
    }
}