3Sum leetcode

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

Note:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2 -1 -4},

    A solution set is:
    (-1, 0, 1)
    (-1, -1, 2)

把数组从0到length - 3 进行遍历 , nums[i]为第一个数 从i + 1 到len -1 用2sum的双指针方法查找 找到一个解之后挪动指针继续查找

要注意数组中的重复问题

遍历的复杂度是O(n) 双指针查找的复杂度是O(n) 所以时间复杂度O(n^2)

public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (nums.length < 3 || nums == null) {
            return res;
        }
        Arrays.sort(nums);
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {//防止出现重复
                continue;
            }
            int left = i + 1;
            int right = nums.length - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum > 0) {
                    right--;
                } else if (sum < 0) {
                    left++;
                } else {
                    List<Integer> tem = new ArrayList<Integer>();
                    tem.add(nums[i]);
                    tem.add(nums[left]);
                    tem.add(nums[right]);
                    res.add(tem);
                    left++;
                    right--;
                    while (left < right && nums[left] == nums[left - 1]) {
                        left++;//防止数组出现重复
                    }
                    while (left < right && nums[right] == nums[right + 1]) {
                        right--;
                    }
                }
            }
        }
        return res;
    }
}