【leetcode】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]
]

题解如下：

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if(nums == null || nums.length < 3) {
            return res;
        }
        //排序(这样那些相同的元素便都排在一起了，方便去重以及指针移动方向的判断)
        Arrays.sort(nums);
        
        //遍历，挑出第一个元素
        for(int i= 0;i < nums.length - 2;i++) {
            if(i==0 || (i > 0 && nums[i - 1] != nums[i])) {
                int sum = 0 - nums[i];
                //找剩下那两个元素
                //定义两个指针，一个从头扫到尾，一个自尾扫到头
                int low = i + 1,high = nums.length - 1;
                while(low < high) {
                    //找到两个元素符合要求了
                    if(nums[low] + nums[high] == sum) {
                        res.add(Arrays.asList(nums[i],nums[low],nums[high]));
                        //去重操作
                        while(low < high && nums[low] == nums[low + 1]) {
                            low++;
                        }
                        while(low < high && nums[high] == nums[high -1]) {
                            high--;
                        }
                        low++;
                        high--;
                    }
                    //如果和小于sum，意味着目标值可能在low指针的右边
                    else if(nums[low] + nums[high] < sum) {
                        low++;
                    } else {
                        high--;
                    }
                }
            }
        }        
        return res;        
    }
}