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> fourSum(vector& nums, int target) {
        int start,end;
        int nSize = nums.size();
        vector triplet;
        vector> triplets;
        set> 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>::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:
2

Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (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& A, vector& B, vector& C, vector& D) {
        int nResult = 0;
        map 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(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;
    }