我可以将数组分区为 K 大小吗？

Question

我正在尝试根据这个问题实现算法：Need idea for solving this algorithm puzzle，但我遗漏了一些边缘情况，这导致我的代码进入无限循环。我可以通过做一些修饰来修复它，但它表明我不理解算法。

谁能帮帮我，我缺少什么？

#include <stdio.h>

#define max(a, b) (((a)>(b))?(a):(b));
int get_max(int *a, int i, int size)
{
    if (i >= size)
        return 0;
    return max(a[i], get_max(a, i+1, size));
}

int get_sum(int *a, int i, int size)
{
    if (i >= size)
        return 0;
    return a[i] + get_sum(a, i+1, size);
}

int get_partition(int *a, int size, int bound) {
    int running_sum = 0;
    int partitions = 0, i;

    for (i=0;i<size;i++) {
        if (a[i] + running_sum <= bound) {
            running_sum += a[i];
        } else {
            running_sum = 0;
            running_sum += a[i];
            partitions++;
        }
    }
    return partitions;
}

int foo(int *a, int size, int k)
{
    int lower = get_max(a, 0, size);
    int higher = get_sum(a, 0, size);
    int partition;

    while (lower < higher) {
        int bound = (lower + (higher))/2;

        partition = get_partition(a, size, bound);
        printf("partition %d bound %d lower %d higher %d\n", partition, bound, lower, higher);
        if (partition >= k) 
            lower = bound;
        else
            higher = bound;
    }
    return partition;
}

#define SIZE(a) sizeof(a)/sizeof(a[0])
int main(void) {
    int a[] = {2, 3, 4, 5, 6};
    printf("%d\n", foo(a, SIZE(a), 3));
    return 0;
}

输出：

partition 1 bound 13 lower 6 higher 20
partition 2 bound 9 lower 6 higher 13
partition 3 bound 7 lower 6 higher 9
partition 3 bound 8 lower 7 higher 9
partition 3 bound 8 lower 8 higher 9
...last line keeps repeating.

Answer 1

你有几个错误：

在二进制搜索期间，您的 while 测试应该是 while (lower+1 < higher) { 而不是 while (lower < higher) {。当 lower = 8、higher = 9 时，您将进入无限循环。在此阶段，您的 bound 将是 (lower+higher)/2=8，您将更新 lower = bound，这不会改变任何内容。
在 foo 的末尾你应该 return higher （不是分区）因为你的二分搜索不变量是为了让 bound <= lower 你可以将数组分区在超过 k 个部分，对于 bound >= higher 个部分，您可以将其分区为 k 个或更少。
你计算的get_partition是错误的。您没有考虑最后一个分区组，因为您只在溢出 running_sum 时更新 partitions。在for循环之后你应该有声明：
```
if (running_sum > 0) 
    partitions++; 
```

综合起来：

#include <stdio.h>

#define max(a, b) (((a)>(b))?(a):(b));
int get_max(int *a, int i, int size)
{
    if (i >= size)
        return 0;
    return max(a[i], get_max(a, i+1, size));
}

int get_sum(int *a, int i, int size)
{
    if (i >= size)
        return 0;
    return a[i] + get_sum(a, i+1, size);
}

int get_partition(int *a, int size, int bound) {
    int running_sum = 0;
    int partitions = 0, i;

    for (i=0;i<size;i++) {
        if (a[i] + running_sum <= bound) {
            running_sum += a[i];
        } else {
            running_sum = 0;
            running_sum += a[i];
            partitions++;
        }
    }
    if (running_sum > 0)
        partitions++;
    return partitions;
}

int foo(int *a, int size, int k)
{
    int lower = get_max(a, 0, size);
    int higher = get_sum(a, 0, size);
    int partition;

    while (lower+1 < higher) {
        int bound = (lower + (higher))/2;

        partition = get_partition(a, size, bound);
        printf("partition %d bound %d lower %d higher %d\n", partition, bound, lower, higher);
        if (partition > k)
            lower = bound;
        else
            higher = bound;
    }
    printf("partition %dlower %d higher %d\n", partition, lower, higher);
    return higher;
}

#define SIZE(a) sizeof(a)/sizeof(a[0])
int main(void) {
    int a[] = {2, 3, 4, 5, 6};
    printf("%d\n", foo(a, SIZE(a), 3));
    return 0;
}

我可以将数组分区为 K 大小吗？

Can I partition the array in K sizes?

algorithm

partitioning

dynamic-programming

greedy