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  • 454. 4Sum II

    package LeetCode_454
    
    /**
     * 454. 4Sum II
     * https://leetcode.com/problems/4sum-ii/
     * 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
     * */
    class Solution {
        /*
        * solution: HashMap, key is sum of A[i] and B[j], value is number of occurrences number of sum,
        * then count the number of -(C[i] and D[j]), for example A[i]+B[j]=1,C[i]+D[j]=-1, so sum up 1,-1 is 0;
        * Time(n^2), Space:O(n^2)
        * */
        fun fourSumCount(A: IntArray, B: IntArray, C: IntArray, D: IntArray): Int {
            val map = HashMap<Int, Int>()
            for (i in A.indices) {
                for (j in B.indices) {
                    val sum = A[i] + B[j]
                    map.put(sum, map.getOrDefault(sum, 0) + 1)
                }
            }
            var result = 0
            for (i in C.indices) {
                for (j in D.indices) {
                    val sum = C[i] + D[j]
                    result += map.getOrDefault(-1 * sum, 0)
                }
            }
            return result
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/14159332.html
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