Java分治法实现最大子数组

问题描述

寻找数组A的和最大的非空连续子数组。例如：数组 A = {13, -3, -25, 20, -3, -16, -23, 18, 20,
-7, 12, -5, -22, 15, -4, 7}的和最大的连续子数组为{18, 20, -7, 12}，最大和为43，所以{18, 20, -7, 12}就是A的最大子数组； 数组{1, -4, 3, -4}的最大子数组为{3}。

解决方案

采用分治策略：将数组分为两个规模相等的子数组，分别求子数组的最大子数组，以及跨越中点的最大子数组，然后将左子数组、右子数组、跨越中点三种情况的最大子数组比较取最大值。

public class MaxmunArray {
    public static void main(String[] args) 
    {
        //int[] arr = {13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7}
        Scanner sc = new Scanner(System.in);
        System.out.println("请输入一串整数并在输入时用英文逗号隔开：");
        String inputString = sc.next().toString();
        String stringArray[] = inputString.split(",");
        int num[] = new int[stringArray.length];
        for (int i = 0; i < stringArray.length; i++) {
            num[i] = Integer.parseInt(stringArray[i]);
        }
        SubArray max_subarray = find_maximum_subarray(num, 0, num.length - 1);
        System.out.println("low:" + max_subarray.low + ", high:" + max_subarray.high + ", sum:" + max_subarray.sum);
    }

    // 查找最大子数组
    private static SubArray find_maximum_subarray(int[] arr, int low, int high) {
        if (low == high) {
            return new SubArray(low, high, arr[low]);
        } else {
            int mid = (low + high) / 2;
            SubArray left = find_maximum_subarray(arr, low, mid);// 递归求左子数组中的最大子数组
            SubArray right = find_maximum_subarray(arr, mid + 1, high);// 递归求右子数组中的最大子数组
            SubArray cross = find_max_crossing_subarray(arr, low, mid, high);// 求跨越中点的最大子数组
            if (left.sum >= right.sum && left.sum >= cross.sum) {
                return left;
            } else if (right.sum >= left.sum && right.sum >= cross.sum) {
                return right;
            } else return cross;
        }
    }

    // 查找包含中点的最大子数组
    private static SubArray find_max_crossing_subarray(int[] arr, int low, int mid, int high) {
        int left_sum = arr[mid];
        int max_left = mid;
        int sum = 0;
        for (int i = mid; i >= low; i--) {// 左边最大和
            sum += arr[i];
            if (sum > left_sum) {
                left_sum = sum;
                max_left = i;
            }
        }
        int right_sum = arr[mid + 1];
        int max_right = mid + 1;
        sum = 0;
        for (int i = mid + 1; i <= high; i++) {// 右边最大和
            sum += arr[i];
            if (sum > right_sum) {
                right_sum = sum;
                max_right = i;
            }
        }
        return new SubArray(max_left, max_right, left_sum + right_sum);
    }

    private static class SubArray {
        int low;
        int high;
        int sum;

        SubArray(int low, int high, int sum) {
            this.low = low;
            this.high = high;
            this.sum = sum;
        }
    }
}

时间复杂度

$$\Theta{(n\log{n})}$$

来源：分治法-最大子数组问题