Java [leetcode 15] 3Sum

问题描述：

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)

解题思路

对于这样的无序数组首先进行升序排列，使之变成非递减数组，调用java自带的Arrays.sort()方法即可。

然后每次固定最小的那个数，在后面的数组中找出另外两个数之和为该数的相反数即可。

具体的过程可参考博客：http://blog.csdn.net/zhouworld16/article/details/16917071

代码实现：

public class Solution {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();

    public List<List<Integer>> threeSum(int[] nums) {
        int length = nums.length;
        if (nums == null || length < 3)
            return ans;
        Arrays.sort(nums);
        for (int i = 0; i < length - 2; ++i) {
            if (i > 0 && nums[i] == nums[i - 1])
                continue;
            findTwoSum(nums, i + 1, length - 1, nums[i]);
        }
        return ans;
    }

    public void findTwoSum(int[] num, int begin, int end, int target) {
        while (begin < end) {
            if (num[begin] + num[end] + target == 0) {
                List<Integer> list = new ArrayList<Integer>();
                list.add(target);
                list.add(num[begin]);
                list.add(num[end]);
                ans.add(list);
                while (begin < end && num[begin + 1] == num[begin])
                    begin++;
                begin++;
                while (begin < end && num[end - 1] == num[end])
                    end--;
                end--;
            } else if (num[begin] + num[end] + target > 0)
                end--;
            else
                begin++;
        }
    }
}