我正在编写一个代码来查找C中的最大和连续子数组。根据我的说法，逻辑似乎很好，但输出仍然不正确。请查看代码。该算法将较大的阵列分成2个子阵列。然后通过检查左数组，右数组以及包含中点的数组来检查最大和子数组（它将从中点检查左右，然后返回包含中点的最大和子数组）。
int* cross_max(int arr[], int low, int mid, int high)
{
int left_max, left_sum = -2000;
int sum = 0;
int i;
for(i=mid; i>=low;i--)
{
sum = sum + arr[i];
if(sum > left_sum)
{
left_sum = sum;
left_max = i;
}
}
int right_max, right_sum = -2000;
for(i=mid+1; i<=high;i++)
{
sum = sum + arr[i];
if(sum > right_sum)
{
right_sum = sum;
right_max = i;
}
}
// 0 - sum
// indices - 1,2
int temp_arr[3] = {0,0,0};
temp_arr[0] = left_sum + right_sum;
temp_arr[1] = left_max;
temp_arr[2] = right_max;
int *p = temp_arr;
printf("\n Maximum sum = %d\n",*p);
printf("\n low = %d\n",*(p+1));
printf("\n high = %d\n",*(p+2));
return p;
}
int* find_max(int arr[], int low, int high)
{
int temp_arr[3] = {0,0,0};
if(low == high)
{
temp_arr[0] = arr[low];
temp_arr[1] = low;
temp_arr[2] = low;
int *q = temp_arr;
return q;
}
int mid = (low + high)/2;
int* a1 = find_max(arr,low,mid);
int* a2 = find_max(arr,mid+1,high);
int* a3 = cross_max(arr,low,mid,high);
if (*a1 > *a2 && *a1 > *a3)
return a1;
else if (*a2 > *a1 && *a2 > *a3)
return a2;
else
return a3;
}
int main()
{
int arr[8] = {1,1,2,-2,3,3,4,-4};
int *point = find_max(arr,0,7);
printf("\n Maximum sum = %d\n",*point);
printf("\n low = %d\n",*(point+1));
printf("\n high = %d\n",*(point+2));
return 0;
}
答案 0 :(得分：6)
稍微偏离主题，但这个问题以解决它的最佳方式而闻名（在线性时间内）。您可以从规范中完全导出代码。
首先，正式定义问题：
给定：整数数组A[0, N)
必填：
max(0 <= p <= q <= N : sum(p, q))
where sum(p, q) = sum(p <= i < q : A[i])
<强>方法：
设X(n) = max(0 <= p <= q <= n : sum(p, q))，然后我们需要找到X(N)。我们通过归纳来做到这一点：
X(0) = max(0 <= p <= q <= 0 : sum(p, q))
= sum(0, 0)
= sum(0 <= i < 0 : A[i])
= 0
和
X(n+1) = max(0 <= p <= q <= n+1 : sum(p, q))
= max(max(0 <= p <= q <= n : sum(p, q)), max(0 <= p <= n+1 : sum(p, n+1)))
= max(X(n), Y(n+1))
其中Y(n) = max(0 <= p <= n : sum(p, n))。我们现在还通过归纳确定Y(n)：
Y(0) = max(0 <= p <= 0 : sum(p, 0))
= sum(0, 0)
= 0
和
Y(n+1) = max(0 <= p <= n+1 : sum(p, n+1))
= max(max(0 <= p <= n : sum(p, n+1)), sum(n+1, n+1)))
= max(max(0 <= p <= n : sum(p, n)) + A[n], 0)
= max(Y(n) + A[n], 0)
<强>代码：
使用上面的分析，代码很简单。
int arr[8] = {1,1,2,-2,3,3,4,-4};
int N = 8;
int x = 0;
int y = 0;
for (int n = 0; n < N; n++) {
y = max(y + arr[n], 0);
x = max(x, y);
}
printf("Maximum sum = %d\n", x);
与
int max(int a, int b) {
if (a > b)
return a;
else
return b;
}
答案 1 :(得分：4)
您的代码中存在一些未定义行为的问题：
首先，您将9作为high传递，它将用于索引八元素数组的第十个元素。这将是第十个，因为cross_max在i <= high时循环，因此您将索引arr[9]。请记住，数组索引从零到大小减去一（因此您可以为您的数组从0索引到7。超出范围的索引将包含未定义（即随机）值。
第二个问题是你从cross_max返回指向局部变量的指针。当您使用返回的指针时，这将导致未定义的行为。局部变量仅在声明它们的范围内有效，并且当函数返回时，局部变量使用的内存区域将被回收并用于下一个函数。
答案 2 :(得分：0)
算法效率不高。时间复杂度为o(n^2)。这是一个动态编程算法，o(n)。
/*************************************************************************
> File Name: subarray.cpp
> Author: luliang
> Mail: lulyon@126.com
> Created Time: 2013/09/10 Tuesday 15:49:23
************************************************************************/
#include <stdio.h>
typedef struct {
int low;
int high;
int sum;
}DPInfoType;
int main()
{
int arr[8] = {1,1,2,-2,3,3,4,-4};
const int n = sizeof(arr) / sizeof(arr[0]);
DPInfoType dp[n];
dp[0].low = 0;
dp[0].high = 0;
dp[0].sum = arr[0];
for(int i = 1; i < n; ++i) {
if(dp[i - 1].sum > 0) {
dp[i].low = dp[i - 1].low;
dp[i].high = i;
dp[i].sum = dp[i - 1].sum + arr[i];
}
else {
dp[i].low = i;
dp[i].high = i;
dp[i].sum = arr[i];
}
}
int max_index = 0;
for(int i = 1; i < n; ++i) {
if(dp[max_index].sum < dp[i].sum) max_index = i;
}
printf("\n Maximum sum = %d\n", dp[max_index].sum);
printf("\n low = %d\n", dp[max_index].low);
printf("\n high = %d\n", dp[max_index].high);
return 0;
}
答案 3 :(得分：0)
如前所述，在您的代码中使用指针是不合适的。 这段代码对我有用。
#include <stdio.h>
#define INF 1000000
int max (int a, int b)
{
if (a < b)
return b;
return a;
}
int findMaxCrossingSubarray (int arr[], int low, int mid, int high, int *start, int *end)
{
int i, left, right;
int max_left, max_right;
int left_sum = -INF;
int sum = 0;
for (i = mid; i >= 0; i--) {
sum += arr[i];
if (sum > left_sum) {
left_sum = sum;
max_left = i;
}
}
int right_sum = -INF;
sum = 0;
for (i = mid + 1; i <= high; i++) {
sum += arr[i];
if (sum > right_sum) {
right_sum = sum;
max_right = i;
}
}
*start = max_left;
*end = max_right;
return left_sum + right_sum;
}
int findMaxSubarray (int arr[], int low, int high, int *start, int *end)
{
if (low == high)
return arr[low];
int mid = (high - low)/2 + low;
int start1, start2, start3;
int end1, end2, end3;
// initialization of start and end for terminal cases.
start1 = start3 = low;
start2 = mid + 1;
end1 = mid;
end2 = end3 = high;
int sum1 = findMaxSubarray(arr, low, mid, &start1, &end1);
int sum2 = findMaxSubarray(arr, mid + 1, high, &start2, &end2);
int sum3 = findMaxCrossingSubarray(arr, low, mid, high, &start3, &end3);
int res = max(max(sum1, sum2), sum3);
if (res == sum1) {
*start = start1;
*end = end1;
}
if (res == sum2) {
*start = start2;
*end = end2;
}
if (res == sum3) {
*start = start3;
*end = end3;
}
return res;
}
int main(int argc, char const *argv[])
{
int size, i, item, result;
printf("Enter the size of array: ");
scanf("%d",&size);
int arr[size];
printf("Enter the array:\n");
for (i = 0; i < size; ++i) {
scanf("%d",&item);
arr[i] = item;
}
int start = 0, end = size-1;
result = findMaxSubarray(arr, 0, size-1, &start, &end);
printf("Result: %d, start: %d and end: %d.\n", result, start, end);
return 0;
}