LeetCode 18. 4Sum and LeetCode 454. 4Sum II

【题目 LeetCode 18】

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: 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]]

【题解】

本题的思路如下： 

1、对数组排序
2、确定四元数中的前两个（a，b）
3、遍历剩余数组确定两外两个（c，d），确定cd时思路跟3Sum确定后两个数据一样，二分查找左右逼近。
4、在去重时采用set集合

【代码】

    vector<vector<int>> fourSum(vector<int>& nums, int target) {        int start,end;        int nSize = nums.size();        vector<int> triplet;        vector<vector<int>> triplets;        set<vector<int>> sets;        sort(nums.begin(),nums.end());        for(int i = 0;i < nSize - 3;i++){            for(int j = i + 1;j < nSize - 2;j++){                //二分查找                start = j + 1;                end = nSize - 1;                while(start < end){                    int curSum = nums[i] + nums[j] + nums[start] + nums[end];                    if(target == curSum){                        triplet.clear();                        triplet.push_back(nums[i]);                        triplet.push_back(nums[j]);                        triplet.push_back(nums[start]);                        triplet.push_back(nums[end]);                        sets.insert(triplet);                        start++;                        end--;                    }                    else if (target > curSum)                        start ++;                    else                        end --;                }            }        }        set<vector<int>>::iterator it;        for(it = sets.begin(); it != sets.end(); it++)            triplets.push_back(*it);                    return triplets;          }

【题目 LeetCode 454】

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:

Input:A = [ 1, 2]B = [-2,-1]C = [-1, 2]D = [ 0, 2]Output:2Explanation:The two tuples are:1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

【题解】
本题的难度相对于LeetCode 18来说，简单了不少，思路如下：

1、先确定前两个数组元素的和，并用map保存key值(和)和value(两个数组元素的和相同的个数)

2、接着计算后两个数组的元素的和的负数，如果和前两个数组元素的和相等，证明四个数组的和值为0

【代码】

    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {        int nResult = 0;        map<int, int> Match;        int nSize = A.size();                for (int i = 0; i < nSize; i++) {            for (int j = 0; j < nSize; j++) {                int nSum = A[i] + B[j];                if (Match.find(nSum) == Match.end())                    Match.insert(pair<int, int>(nSum, 0));                                    Match[nSum] += 1;            }        }                for (int i = 0; i < nSize; i++) {            for (int j = 0; j < nSize; j++) {                int nSum = -(C[i] + D[j]);                if (Match.find(nSum) != Match.end())                    nResult += Match[nSum];            }        }                return nResult;    }