LeetCode 15 - Longest Common Prefix

最新推荐文章于 2024-06-09 16:29:53 发布

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最新推荐文章于 2024-06-09 16:29:53 发布

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分类专栏：编程题目文章标签： C++ LeetCode

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本文介绍了一种解决寻找数组中三个数之和为零的问题的方法，包括算法实现、步骤和实例演示。

3Sum

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)

My Code

class Solution {
public:
    vector<vector<int> > threeSum(vector<int>& nums) {
        vector<vector<int> > results;

        int size = nums.size();
        if (size < 3)
            return results;

        sort(nums.begin(), nums.end());
        int zero_idx = -1;
        for (int i = 0; i < size; i++)
        {
            if (nums[i] == 0)
            {
                zero_idx = i;
                break;
            }
            else if (nums[i] > 0)
                break;
        }

        int min_val = nums.front(), max_val = nums.back();
        if (min_val > 0 || max_val < 0)
            return results;

        int last_min_val = min_val - 1;
        for (int k = 0; k < size - 2; k++)
        {
            if (nums[k] == last_min_val)
                continue;

            if (nums[k] > 0)
                break;

            int last_mid_val = nums[k] - 1;
            int target_val = 0 - nums[k];
            for (int i = k + 1; i < size - 1; i++)
            {
                if (nums[i] == last_mid_val)
                    continue;

                if (nums[i] + nums[i+1] > target_val)
                    break;

                int last_max_val = last_mid_val - 1;
                for (int j = i + 1; j < size; j++)
                {
                    if (nums[j] == last_max_val)
                        continue;

                    if (nums[i] + nums[j] == target_val)
                    {
                        vector<int> result(3, 0);
                        result[0] = nums[k];
                        result[1] = nums[i];
                        result[2] = nums[j];
                        results.push_back(result);
                        break;
                    }
                    else if (nums[i] + nums[j] > target_val)
                        break;

                    last_max_val = nums[j];
                }

                last_mid_val = nums[i];
            }

            last_min_val = nums[k];
        }

        return results;
    }
};

Runtime: 64 ms