技術標籤：JavaC++

問題：

難度：medium

說明：

給出四個陣列，將四個陣列內拿一個元素，返回四個相加 == 0 的組合個數。

題目連線：https://leetcode.com/problems/4sum-ii/submissions/

輸入範圍：

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 -228to 228- 1 and the result is guaranteed to be at most 231- 1.

輸入案例：

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

我的程式碼：

這個也只是化問題為 2Sum，把集合兩兩相加即可。使用 map 也挺耗時，因為500 * 500 * 2 長度陣列的時間和空間複雜度，沒有特別好的辦法，最多弄成陣列可以節省點空間。

java-map：

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        int len = A.length, count = 0;
        HashMap<Integer,Integer> mapA = new HashMap<Integer,Integer>();
        HashMap<Integer,Integer> mapB = new HashMap<Integer,Integer>();
        for(int i = 0;i < len;i ++) {
            for(int j = 0;j < len;j ++) {
                int a = A[i] + B[j];
                int b = - C[i] - D[j];
                mapA.put(a, mapA.getOrDefault(a, 0) + 1);
                mapB.put(b, mapB.getOrDefault(b, 0) + 1);
            }
        }
        for(int value : mapA.keySet()) 
            if(mapB.containsKey(value)) count += mapA.get(value) * mapB.get(value);
        return count;
    }
}
 

java-arr：

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        int len = A.length, count = 0, times = len * len;
        int AB[] = new int[times];
        int CD[] = new int[times];
        for(int i = 0;i < len;i ++) {
            int slot = i * len;
            for(int j = 0;j < len;j ++) {
                AB[slot + j] = A[i] + B[j];
                CD[slot + j] = - C[i] - D[j];
            }
        }
        Arrays.sort(AB);
        Arrays.sort(CD);
        int i = 0,j = 0;
        A:for(;i < times;) {
            for(;j < times;j ++) {
                if(AB[i] == CD[j]) {
                    int ABT = 1, CDT = 1;
                    while(++ i < times && AB[i] == AB[i - 1]) ABT ++;
                    while(++ j < times && CD[j] == CD[j - 1]) CDT ++;
                    count += ABT * CDT;
                    continue A;
                } else if(AB[i] < CD[j]) break;
            } i ++;
        }
        return count;
    }
}

c++ arr：

class Solution {
public:
	int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
		int count = 0, len = A.size(), times = len * len;
		int *AB = new int[times], *CD = new int[times];
		for (int i = 0; i < len; i++) {
			int slot = i * len;
			for (int j = 0; j < len; j++) {
				int site = slot + j;
				AB[site] = A[i] + B[j];
				CD[site] = -C[i] - D[j];
			}
		}
		sort(AB, AB + times);
		sort(CD, CD + times);
		int i = 0, j = 0;
		A:for (; i < times; ) {
			for (; j < times; j++) {
				if (AB[i] == CD[j]) {
					int ABT = 1, CDT = 1;
					while (++i < times && AB[i] == AB[i - 1]) ABT++;
					while (++j < times && CD[j] == CD[j - 1]) CDT++;
					count += ABT * CDT;
					goto A;
				} else if(AB[i] < CD[j]) break;
			} i++;
		}
		return count;
	}
};