LeetCode 3Sum

最新推荐文章于 2022-01-27 22:44:32 发布

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本文介绍了一种高效算法来寻找数组中所有和为零的唯一三元组，包括算法实现和示例代码。

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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;
    }
}