LeetCode Top Interview Questions 454. 4Sum II (Java版; Meidum)

浅浅的花香味﹌ 2023-07-03 02:58 19阅读 0赞

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## LeetCode Top Interview Questions 454. 4Sum II (Java版; Meidum) ##

#### 题目描述 ####

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

#### 第一次做; 把下面的写法稍微整理一下, 看起来更简洁; 核心:把four sum转成two sum ####

class Solution { 
        public int fourSumCount(int[] A, int[] B, int[] C, int[] D) { 
            HashMap<Integer, Integer> sumAB = new HashMap<>();
            for(int a: A){ 
                for(int b: B){ 
                    int sum = a+b;
                    sumAB.put(sum, sumAB.getOrDefault(sum,0)+1);
                }
            }
            
            int count=0; 
            for(int c:C){ 
                for(int d:D){ 
                    int cur = -c-d;
                    if(sumAB.containsKey(cur))
                        count+=sumAB.get(cur);
                }
            }
            return count;
        }
    }

#### 第一次做; 两个两个的处理, 相当于将four sum转成了two sum; 时间复杂度O(N^2), 空间复杂度O(N^2) ####

//两个两个弄, 相当于转换成了two sum问题
    class Solution { 
        public int fourSumCount(int[] A, int[] B, int[] C, int[] D) { 
            //key是ab的求和结果, value是这个结果出现的次数
            HashMap<Integer, Integer> ab = new HashMap<>();
            
            int count=0;
            
            for(int a=0; a<A.length; a++){ 
                for(int b=0; b<B.length; b++){ 
                    int sum = A[a] + B[b];
                    ab.put(sum, ab.getOrDefault(sum, 0)+1);
                }
            }
            for(int c=0; c<C.length; c++){ 
                for(int d=0; d<D.length; d++){ 
                    int sum = C[c] + D[d];
                    int cur = - sum;
                    if(ab.containsKey(cur))
                        count += ab.get(cur);
                }
            }
            return count;
        }
    }

#### 第一次做; 暴力; 超时46/48; 需要用空间换时间; 时间复杂度O(N^4), 空间复杂度O(1) ####

//暴力, 四层循环
    class Solution { 
        public int fourSumCount(int[] A, int[] B, int[] C, int[] D) { 
            int count = 0;
            for(int a=0; a<A.length; a++){ 
                for(int b=0; b<B.length; b++){ 
                    for(int c=0; c<C.length; c++){ 
                        for(int d=0; d<D.length; d++){ 
                            if(A[a]+B[b]+C[c]+D[d]==0)
                                count++;
                        }
                    }
                }
            }
            return count;
        }
    }

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