15. 3Sum - medium

注意要跳过重复。
把每个element 看成三个数中的一个，然后在这个数后面剩余的数里，从最大和最小的数开始做2sum.

public class Solution {
    public List> threeSum(int[] nums) {
        Arrays.sort(nums);
        List> ans = new ArrayList>();
        
        for (int i=0; i0 && nums[i] != nums[i-1])) {
                int lo = i+1, hi = nums.lengh-1, sum = 0-nums[i];
                // try sum:
                while (lo < hi) {
                    if (nums[lo] + num[hi] == sum) {
                        res.add(Arrays.asList(num[i], num[lo], num[hi]));
                        // skip same elements:
                        while (lo < hi && nums[lo] == nums[lo+1]) lo++;
                        while (lo < hi && nums[hi] == nums[hi-1]) hi--;
                        lo++; hi--;
                    }
                    else if (num[lo] + num[hi] < sum) lo++;
                    else  hi--;
                }
            }
        }
        return res;
    }
}