leetcode 18.4Sum

1.题目

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.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    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)

从一个数组中找到四个数，使这四个数的和为target，返回这四个数。

2.思路 

思路还是跟3sum 一样，先贴上我的代码，有很多的地方可以优化。

class Solution { //192mspublic:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> ans;        int len=nums.size();        if(len<4) return {};        sort(nums.begin(),nums.end());        for(int i=0;i<len-3;)        {            for(int j=i+1;j<len-2;)            {                int head=j+1,tail=len-1;                while(head<tail)                {                    if(nums[head]+nums[tail]+nums[i]+nums[j] == target)                    {    ans.push_back(vector<int>{nums[i],nums[j],nums[head],nums[tail]});                         do{tail--;} while(nums[tail] == nums[tail+1] && head<tail);                         do{head++;} while(nums[head] == nums[head-1] && head<tail);                    }                    else if(nums[head]+nums[tail]+nums[i]+nums[j] > target)                    {                        do{tail--;} while(nums[tail] == nums[tail+1] && head<tail);                    }                    else                    {                        do{head++;} while(nums[head] == nums[head-1] && head<tail);                    }                }                do{j++;} while(nums[j] == nums[j-1] && j<len-2);                        }                do{i++;} while(nums[i] == nums[i-1] && i<len-3);        }        return ans;    }};

再贴上disguss中的高票答案，https://leetcode.com/discuss/67417/my-16ms-c-code

class Solution { //16mspublic:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> total;        int n = nums.size();        if(n<4)  return total;        sort(nums.begin(),nums.end());        for(int i=0;i<n-3;i++)        {            if(i>0&&nums[i]==nums[i-1]) continue; //这里，相等的数不重复检索            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;  //如果初始值都大于target，那么之后的肯定找不到了            if(nums[i]+nums[n-3]+nums[n-2]+nums[n-1]<target) continue;//在i固定的情况下，加上最大的三个数都小于target            for(int j=i+1;j<n-2;j++)            {                if(j>i+1&&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[n-2]+nums[n-1]<target) continue;                int left=j+1,right=n-1;                while(left<right){                    int sum=nums[left]+nums[right]+nums[i]+nums[j];                    if(sum<target) left++;                    else if(sum>target) right--;                    else{                        total.push_back(vector<int>{nums[i],nums[j],nums[left],nums[right]});                        do{left++;}while(nums[left]==nums[left-1]&&left<right);                        do{right--;}while(nums[right]==nums[right+1]&&left<right);                    }                }            }        }        return total;    }};