分治法-最大子数组问题

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}。

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。

Java代码实现：

class FindMaximumSubArray
{
public static void main(String[] args)
{
int[] arr = {13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7};
SubArray max_subarray = find_maximum_subarray(arr, 0, arr.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;
}
}

}
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该算法的运行时间为Θ(nlgn)。
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