zoukankan      html  css  js  c++  java
  • [LeetCode] 454. 4Sum II 四数之和之二

    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

    这道题是之前那道 4Sum 的延伸,让我们在四个数组中各取一个数字,使其和为0。那么坠傻的方法就是遍历所有的情况,时间复杂度为 O(n4)。但是既然 Two Sum 那道都能将时间复杂度缩小一倍,那么这道题使用 HashMap 是否也能将时间复杂度降到 O(n2) 呢?答案是肯定的,如果把A和B的两两之和都求出来,在 HashMap 中建立两数之和跟其出现次数之间的映射,那么再遍历C和D中任意两个数之和,只要看哈希表存不存在这两数之和的相反数就行了,参见代码如下:

    解法一:

    class Solution {
    public:
        int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
            int res = 0;
            unordered_map<int, int> m;
            for (int i = 0; i < A.size(); ++i) {
                for (int j = 0; j < B.size(); ++j) {
                    ++m[A[i] + B[j]];
                }
            }
            for (int i = 0; i < C.size(); ++i) {
                for (int j = 0; j < D.size(); ++j) {
                    int target = -1 * (C[i] + D[j]);
                    res += m[target];
                }
            }
            return res;
        }
    };

    下面这种方法用了两个 HashMap 分别记录 AB 和 CB 的两两之和出现次数,然后遍历其中一个 HashMap,并在另一个 HashMap 中找和的相反数出现的次数,参见代码如下:

    解法二:

    class Solution {
    public:
        int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
            int res = 0, n = A.size();
            unordered_map<int, int> m1, m2;
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    ++m1[A[i] + B[j]];
                    ++m2[C[i] + D[j]];
                }
            }
            for (auto a : m1) res += a.second * m2[-a.first];
            return res;
        }
    };

    类似题目:

    4Sum

    参考资料:

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

    https://leetcode.com/problems/4sum-ii/discuss/93920/Clean-java-solution-O(n2)

    https://leetcode.com/problems/4sum-ii/discuss/93925/Concise-C%2B%2B-11-code-beat-99.5

    LeetCode All in One 题目讲解汇总(持续更新中...)

  • 相关阅读:
    Maven依赖
    Maven坐标
    初识Maven POM
    Maven配置
    相似文本文档分析之SimHash算法
    Ubuntu14.10下JDK编译安装详细操作说明
    Ubuntu14.10下Tomcat编译安装 详细操作说明
    ubuntu 14.10 编译安装 Ruby
    ubuntu 14.10 编译安装 Python
    ubuntu 14.10 编译安装 Golang
  • 原文地址:https://www.cnblogs.com/grandyang/p/6073317.html
Copyright © 2011-2022 走看看