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Legal Studies用 C++ 中的分治法找出最大和连续子数组
假设我们有一个包含正值和负值的数据列表。我们需要求和一个连续子数组，其和为最大值。假设该列表包含 {-2, -5, 6, -2, -3, 1, 5, -6}，则最大子数组之和为 7。即 {6, -2, -3, 1, 5} 之和
我们将使用分治法来解决此问题。步骤如下所示 −
步骤 −
- 将数组分为两部分
- 找到以下三者的最大值
- 左子数组的最大子数组和
- 右子数组的最大子数组和
- 子数组的最大子数组和，其中子数组穿过中点
示例
#include <iostream>
using namespace std;
int max(int a, int b) {
return (a > b)? a : b;
}
int max(int a, int b, int c) {
return max(max(a, b), c);
}
int getMaxCrossingSum(int arr[], int l, int m, int h) {
int sum = 0;
int left = INT_MIN;
for (int i = m; i >= l; i--) {
sum = sum + arr[i];
if (sum > left)
left = sum;
}
sum = 0;
int right = INT_MIN;
for (int i = m+1; i <= h; i++) {
sum = sum + arr[i];
if (sum > right)
right = sum;
}
return left + right;
}
int maxSubArraySum(int arr[], int low, int high) {
if (low == high)
return arr[low];
int mid = (low + high)/2;
return max(maxSubArraySum(arr, low, mid), maxSubArraySum(arr, mid+1, high), getMaxCrossingSum(arr, low, mid, high));
}
int main() {
int arr[] = {-2, -5, 6, -2, -3, 1, 5, -6};
int n = sizeof(arr)/sizeof(arr[0]);
int max_sum = maxSubArraySum(arr, 0, n-1);
printf("Maximum contiguous sum is %d", max_sum);
}输出
Valid String
更新时间： 2019-10-21
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