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1 | 1 | /**
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2 |
| - * Finds the maximum subarray using the divide and conquer algorithm |
| 2 | + * Finds the maximum subarray using the divide and conquer algorithm |
3 | 3 | * by Bentley, Jon (1984) (complexity O(n(logn)));
|
4 | 4 | */
|
5 | 5 |
|
6 |
| -/** |
7 |
| - * Accepts an array and range. Finds the maximum sum of elements |
8 |
| - * around the middle of the range. |
9 |
| - * |
10 |
| - * @param {array} array |
11 |
| - * @param {number} left - the left interval of the range |
12 |
| - * @param {number} middle - the middle of the range |
13 |
| - * @param {number} right - the right side of the range |
14 |
| - * @return {number} the maximum sum including the middle element |
15 |
| - */ |
16 |
| -function crossSubarray(array, left, middle, right) { |
17 |
| - var leftSum = -Infinity, |
18 |
| - rightSum = -Infinity, |
19 |
| - sum = 0, |
20 |
| - i; |
| 6 | +(function (exports) { |
| 7 | + |
| 8 | + 'use strict'; |
21 | 9 |
|
22 |
| - for (i = middle; i >= left; i -= 1) { |
23 |
| - if (sum + array[i] >= leftSum) { |
24 |
| - leftSum = sum + array[i]; |
| 10 | + /** |
| 11 | + * Accepts an array and range. Finds the maximum sum of elements |
| 12 | + * around the middle of the range. |
| 13 | + * |
| 14 | + * @param {array} array |
| 15 | + * @param {number} left - the left interval of the range |
| 16 | + * @param {number} middle - the middle of the range |
| 17 | + * @param {number} right - the right side of the range |
| 18 | + * @return {number} the maximum sum including the middle element |
| 19 | + */ |
| 20 | + function crossSubarray(array, left, middle, right) { |
| 21 | + var leftSum = -Infinity, |
| 22 | + rightSum = -Infinity, |
| 23 | + sum = 0, |
| 24 | + i; |
| 25 | + |
| 26 | + for (i = middle; i >= left; i -= 1) { |
| 27 | + if (sum + array[i] >= leftSum) { |
| 28 | + leftSum = sum + array[i]; |
| 29 | + } |
| 30 | + sum += array[i]; |
| 31 | + } |
| 32 | + sum = 0; |
| 33 | + for (i = middle + 1; i < right; i += 1) { |
| 34 | + if (sum + array[i] >= rightSum) { |
| 35 | + rightSum = sum + array[i]; |
| 36 | + } |
| 37 | + sum += array[i]; |
25 | 38 | }
|
26 |
| - sum += array[i]; |
| 39 | + return leftSum + rightSum; |
27 | 40 | }
|
28 |
| - sum = 0; |
29 |
| - for (i = middle + 1; i < right; i += 1) { |
30 |
| - if (sum + array[i] >= rightSum) { |
31 |
| - rightSum = sum + array[i]; |
| 41 | + |
| 42 | + /** |
| 43 | + * Using divide and conquer finds the maximum sum of subarray of the given |
| 44 | + * |
| 45 | + * @param {array} array |
| 46 | + * @param {number} left side of the range |
| 47 | + * @param {number} the right side of the range |
| 48 | + * @return {number} the maximum sum of the elements of |
| 49 | + * subarray whithin the given range |
| 50 | + */ |
| 51 | + function maxSubarrayPartitioner(array, left, right) { |
| 52 | + if (right - left <= 1) { |
| 53 | + return array[left]; |
32 | 54 | }
|
33 |
| - sum += array[i]; |
| 55 | + var middle = Math.floor((left + right) / 2), |
| 56 | + leftSum = maxSubarrayPartitioner(array, left, middle), |
| 57 | + rightSum = maxSubarrayPartitioner(array, middle, right), |
| 58 | + crossSum = crossSubarray(array, left, middle, right); |
| 59 | + |
| 60 | + return Math.max(crossSum, leftSum, rightSum); |
34 | 61 | }
|
35 |
| - return leftSum + rightSum; |
36 |
| -} |
37 | 62 |
|
38 |
| -/** |
39 |
| - * Using divide and conquer finds the maximum sum of subarray of the given |
40 |
| - * |
41 |
| - * @param {array} array |
42 |
| - * @param {number} left side of the range |
43 |
| - * @param {number} the right side of the range |
44 |
| - * @return {number} the maximum sum of the elements of |
45 |
| - * subarray whithin the given range |
46 |
| - */ |
47 |
| -function maxSubarrayPartitioner(array, left, right) { |
48 |
| - if (right - left <= 1) { |
49 |
| - return array[left]; |
| 63 | + /** |
| 64 | + * Returns the maximum sum of the elements of a subarray of the given array |
| 65 | + * |
| 66 | + * @param {array} the array |
| 67 | + * @return the maximum sum |
| 68 | + */ |
| 69 | + function maxSubarray(array) { |
| 70 | + return maxSubarrayPartitioner(array, 0, array.length); |
50 | 71 | }
|
51 |
| - var middle = Math.floor((left + right) / 2), |
52 |
| - leftSum = maxSubarrayPartitioner(array, left, middle), |
53 |
| - rightSum = maxSubarrayPartitioner(array, middle, right), |
54 |
| - crossSum = crossSubarray(array, left, middle, right); |
55 | 72 |
|
56 |
| - return Math.max(crossSum, leftSum, rightSum); |
57 |
| -} |
| 73 | + exports.maxSubarray = maxSubarray; |
58 | 74 |
|
59 |
| -/** |
60 |
| - * Returns the maximum sum of the elements of a subarray of the given array |
61 |
| - * |
62 |
| - * @param {array} the array |
63 |
| - * @return the maximum sum |
64 |
| - */ |
65 |
| -function maxSubarray(array) { |
66 |
| - return maxSubarrayPartitioner(array, 0, array.length); |
67 |
| -} |
| 75 | +}(typeof exports === 'undefined' ? window : exports)); |
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