[leetcode-18-4Sum]

最新推荐文章于 2022-02-10 12:14:07 发布

转载最新推荐文章于 2022-02-10 12:14:07 发布 · 44 阅读

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原文链接：http://www.cnblogs.com/hellowooorld/p/6531103.html

本文详细解析了四数之和问题的算法实现过程，针对给定数组寻找四个元素组合达到特定目标值的所有唯一四元组。通过双指针技巧和排序优化减少搜索空间，避免重复解。

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Given an array S of n integers, are there elements a, b, c,
and d in S such that a + b + c + d = target? Find all unique
quadruplets in the array which gives the sum of target.
For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.
A solution set is:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]

思路：

类似3sum求和，然后在过程中及时约束，判断是否可以提前跳出循环来优化。

vector<vector<int>> fourSum(vector<int>& nums, int target)
    {
        vector<vector<int>>result;
        const int size = nums.size();
        if (size < 4) return result;
        sort(nums.begin(), nums.end());
        vector<int>temp;
        for (int i = 0; i < size - 3; i++)
        {
            if (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) break;//及时跳出循环
            if (nums[i] + nums[size - 3] + nums[size - 2] + nums[size - 1] < target)continue;
            
            for (int j = i + 1; j < size-2; j++)
            {
                if (nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) break;//及时跳出循环
                if (nums[i] + nums[j] + nums[size - 2] + nums[size - 1] < target)continue;
                 
                int left = j + 1;
                int right = size - 1;
                while (left < right)
                {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum < target)left++;
                    else if (sum > target) right--;
                    else
                    {
                        temp = {nums[i],nums[j],nums[left],nums[right]};//新学一招，类似数组赋值
                        result.push_back(temp);
                        while (left < right && nums[left] == temp[2])left++;
                        while (left < right && nums[right] == temp[3])right--;
                    }         
                }
                while (j + 1 < size && nums[j + 1] == nums[j])j++;
            }
            while (i + 1 < size && nums[i + 1] == nums[i])i++;
        }
        return result;
    }

参考：

http://cs-cjl.com/2016/07_04_leetcode_18_4sum

转载于:https://www.cnblogs.com/hellowooorld/p/6531103.html