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454. 4Sum II

时间:2018-01-07 14:26:03      阅读:123      评论:0      收藏:0      [点我收藏+]

标签:var   ref   lin   xpl   ons   bit   +=   为知笔记   make   

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

给定四个列表A,B,C,D的整数值,计算有多少个元组(i,j,k,l),使得A [i] + B [j] + C [k] + D [1]是零。
为了使问题更容易,所有A,B,C,D具有相同的长度N,其中0≤N≤500。所有整数在-228到228-1的范围内,结果保证最多为231 - 1。
  1. /**
  2. * @param {number[]} A
  3. * @param {number[]} B
  4. * @param {number[]} C
  5. * @param {number[]} D
  6. * @return {number}
  7. */
  8. var fourSumCount = function (A, B, C, D) {
  9. let res = 0;
  10. let m = {};
  11. for (let i in A) {
  12. for (let j in B) {
  13. let value = A[i] + B[j];
  14. m[value] = m[value] ? ++m[value] : 1;
  15. }
  16. }
  17. for (let i in C) {
  18. for (let j in D) {
  19. let negativeValue = -1 * (C[i] + D[j]);
  20. if (m[negativeValue]) {
  21. res += m[negativeValue];
  22. }
  23. }
  24. }
  25. return res;
  26. };
  27. let A = [1, 2];
  28. let B = [-2, -1];
  29. let C = [-1, 2];
  30. let D = [0, 2];
  31. let res = fourSumCount(A, B, C, D);
  32. console.log(res);





454. 4Sum II

标签:var   ref   lin   xpl   ons   bit   +=   为知笔记   make   

原文地址:https://www.cnblogs.com/xiejunzhao/p/8227464.html

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