c - Leetcode 算法计算时间过长

Question

https://leetcode.com/problems/pizza-with-3n-slices/

我了解编程的基础知识，最近决定研究 leetcode，所以我很想知道我在这些难题上的立场。警告的话，它看起来并不漂亮。该代码有效，但计算您添加的更多切片需要成倍增加的时间。我的解决方案是可以挽救的还是死路一条，我应该从不同的思路开始？

我的算法的本质适用于这个快速图表，请记住我为此使用 C。第一个循环遍历数组的每个单独元素，它调用一个函数来删除 3 个切片，将它们替换为 -1 作为一种标志。这个函数是递归的，用数组的旧副本从下一个元素中删除切片，直到它遍历每个元素。它将退出第二个循环并从第一个循环的下一个元素重新开始。我想我可以把它全部做成一个递归循环，但我不确定我是否可以编辑 leetcode 的参数来完成他们给我的功能。

int copyArray(int * from, int * to, int size){
    for(int i = 0; i < size; i++)
        to[i] = from[i];
    return 1;
}
int findFirstElement(int *slices, int size){
for(int i = 0; i< size; i++)
    if (slices[i] != -1)
        return i;
}
int findLastElement(int *slices, int size){
for(int i = size-1; i>=0; i--)
    if (slices[i] != -1)
        return i;
        
 
}
int getMax(int *a, int b){
    return (*a>b)? *a: b;
}
 
int removeSlices(int *slices, int pos, int slicesSize, int *max, int size){  //int size is 
//the size of the pizza which will never change unlike
// slicesSize which is decremented by 3 in each level we enter
    int *tempmax;
    int l = *max;
    tempmax = &l;
    int newSlices[size];
    int k = 0;
    int m = 0;
    int j = 0;
    int *maxmax;
    int p = 0;
    maxmax = &p;
    int newSlicesSize = slicesSize;
    copyArray(slices, newSlices, size);
    int firstElement = findFirstElement(newSlices, size);
    int lastElement = findLastElement(newSlices, size);
    *max += newSlices[pos];
    if(pos == firstElement){ // represent the array as a circle 
        newSlices[lastElement] = -1;
        newSlices[pos] = -1;
        newSlices[pos+1] = -1;
    }
    else if(pos == lastElement)
    {
        newSlices[pos-1] = -1;
        newSlices[pos] = -1;
        newSlices[firstElement] = -1;
 
    }
    else {
            newSlices[pos-1] = -1;
            newSlices[pos] = -1;
            newSlices[pos+1] = -1;
    }
 //   for(int i = 0; i<size; i++)
  //      printf("%d,", newSlices[i]);
 //   printf("\n");
    newSlicesSize -= 3;
    if(newSlicesSize == 0)
        return *max;
 
 
    while(k<newSlicesSize){ //the second loop 
    m = findFirstElement(newSlices, size); 
    *tempmax = *max; // yet another max var (sorry), before we enter a new level we have to keep hold of the max so we can return to it once it has finished 
    j = removeSlices(newSlices, m+k, newSlicesSize, tempmax, size);
 
     *maxmax =getMax(maxmax, j); // maxmax holds the max of all decisions/possibilites for only the individual elements of the first loop, not the final max of all elements of the first loop
    k++;
    }
    return *maxmax;
 
}
 
 
 
int maxSizeSlices(int* slices, int slicesSize){
    int *max;
    int k = 0;
    max = &k;
    int j = 0;
    int *maxmax;
    int p = 0;
    maxmax = &p;
    int i = 0;
    while(i < slicesSize){
            *max =0;
            j = removeSlices(slices,i, slicesSize, max,slicesSize);
            *maxmax =getMax(maxmax, j);
            i++;
    }
    return *maxmax;
 
}
int main()
{
   // int slices[] = {6,3,1,2,6,2,4,3,10,4,1,4,6,5,5,3,4,7,6,5,8,7,3,8,8,1,7,1,7,8}; // gave up trying to compute this after 20min
     int slices[] = {1,2,3,4,5,6}; // computes quickly
    int size = 6;
    int max = maxSizeSlices(slices, size);
    printf("final max: %d",max);
    return 0;
}'''

标签： c