LeetCode 454 4Sum II (二分)

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

题目链接：https://leetcode.com/problems/4sum-ii/description/

题目分析：预处理出AB间，CD间任意两个数字的和，对AB预处理出的数组排序，CD中每个值通过二分查找其在AB中相反数的个数，LowerBound函数返回第一个小于等于待查找的数的位置，时间复杂度O(n^2logn)

class Solution {
    
    public int LowerBound(int val, int[] a, int n) {
        int l = 0, r = n - 1, mid;
        while (l <= r) {
            mid = (l + r) >> 1;
            if (a[mid] >= val) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
    
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        int n = A.length, N = n * n;
        int[] AB = new int[N];
        int[] CD = new int[N];
        int cntAB = 0, cntCD = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                AB[cntAB++] = A[i] + B[j];
                CD[cntCD++] = C[i] + D[j];
            }
        }
        Arrays.sort(AB);
        int ans = 0;
        for (int i = 0; i < cntCD; i++) {
            ans += LowerBound(-CD[i] + 1, AB, cntAB) - LowerBound(-CD[i], AB, cntAB);
        }
        return ans;
    }
}