LeetCode 3Sum

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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)

题意：给出一个数组，求所有三个数之和为0的组合

与LeetCode 4Sum(backtrace+剪枝)相似

代码如下：

class Solution {    private void kSum(int[] nums, int start, int k, int target, List<Integer> arr, List<List<Integer>> ans)    {        int len = nums.length;        if (start >= len || 0 == k) return;        if (1 == k)        {            for (int i = start; i < len; i++)            {                if (nums[i] == target)                {                    arr.add(nums[i]);                    ans.add(new ArrayList<Integer>(arr));                    arr.remove(arr.size() - 1);                }            }            return;        }        if (2 == k)        {            int end = len - 1;            while (start < end)            {                int sum = nums[start] + nums[end];                if (sum == target)                {                    arr.add(nums[start]);                    arr.add(nums[end]);                    ans.add(new ArrayList<Integer>(arr));                    arr.remove(arr.size() - 1);                    arr.remove(arr.size() - 1);                    while (start < end && nums[start] == nums[start + 1]) start++;                    start++;                    while (start < end && nums[end] == nums[end - 1]) --end;                    --end;                }                else if (sum < target)                {                    while (start < end && nums[start] == nums[start + 1]) start++;                    start++;                }                else                {                    while (start < end && nums[end] == nums[end - 1]) end--;                    end--;                }            }            return;        }        if (k * nums[start] > target || k * nums[len - 1] < target) return;        for (int i = start; i <= (len - k); i++)        {            if (i > start && nums[i] == nums[i - 1]) continue;            if (k * nums[i] <= target)            {                arr.add(nums[i]);                kSum(nums, i + 1, k - 1, target - nums[i], arr, ans);                arr.remove(arr.size() - 1);            }        }    }    public List<List<Integer>> threeSum(int[] nums)    {        List<List<Integer>> ans = new ArrayList<List<Integer>>();        List<Integer> arr = new ArrayList<Integer>();        Arrays.sort(nums);        kSum(nums, 0, 3, 0, arr, ans);        return ans;    }}

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