leetcode 18 4Sum

来源：互联网发布：手机淘宝网手机淘宝编辑：程序博客网时间：2024/06/06 13:07

问题

https://leetcode.com/problems/4sum/

解法一(138ms)

枚举第一个和第四个数，第二个和第三个数tow pointer
复杂度O(n3)

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> ret;        vector<int> res(4, 0);        if (nums.size() <4) return ret;        sort(nums.begin(), nums.end());        // 注意size_t 类型为unsigned int 小于0 时会出错        for (int i=0; i<(int)nums.size() -3; i++)        {            if (i-1>=0 && nums[i]==nums[i-1])                continue;            for (int j =nums.size()-1; i+3 <= j; j--)            {                if (j+1 < nums.size() && nums[j] == nums[j+1])                    continue;                int k= i+1;                int s = j-1;                while(k< s)                {                    if (k-1 > i && nums[k]==nums[k-1])                    {                        k++;                        continue;                    }                    if (s+1 < j && nums[s] == nums[s+1])                    {                        s--;                        continue;                    }                    int now = nums[i]+ nums[k]+nums[s]+nums[j];                    if (now == target)                    {                        res[0] = nums[i];                        res[1] = nums[k];                        res[2] = nums[s];                        res[3] = nums[j];                        ret.push_back(res);                        k++;                        s--;                    }else if (now < target)                            k++;                        else                            s--;                }            }        }        return ret;    }};

解法二（20ms）
加上剔除。
分别从前往后枚举前两个数，后两个数用tow pointer
这样可以将算法推广到mSum，复杂度为O(nm−1)

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> ret;        vector<int> res(4, 0);        if (nums.size() <4) return ret;        sort(nums.begin(), nums.end());        for (int i=0; i<nums.size() -3; i++)        {            if (i-1>=0 && nums[i]==nums[i-1])                continue;            if (nums[i] + nums[i+1] + nums[i+2] + nums[i+3] > target)                break;            if (nums[i] + nums[nums.size()-1]  + nums[nums.size()-2] + nums[nums.size()-3] < target)                continue;            for (int j =i+1; j  < nums.size() -2; j++)            {                if (j-1 >i && nums[j] == nums[j-1])                    continue;                if (nums[i] + nums[j] + nums[j+1] +nums[j+2] >target)                    break;                if (nums[i] + nums[j] + nums[nums.size() -1] + nums[nums.size()-2] < target)                    continue;                int k= j+1;                int s = nums.size() -1;                while(k< s)                {                    if (k-1 > j && nums[k]==nums[k-1])                    {                        k++;                        continue;                    }                    if (s+1 < nums.size() && nums[s] == nums[s+1])                    {                        s--;                        continue;                    }                    int now = nums[i]+ nums[j]+nums[k]+nums[s];                    if (now == target)                    {                        res[0] = nums[i];                        res[1] = nums[j];                        res[2] = nums[k];                        res[3] = nums[s];                        ret.push_back(res);                        k++;                        s--;                    }else if (now < target)                            k++;                        else                            s--;                }            }        }        return ret;    }};

0 0