虽然工作三年了，但是数据结构还是一塌糊涂，上一份工作就是因为代码写少了所以被开除了，胡玩了三个月之后现在又开始捡原来的知识。
回到了PTA，真怀念上学的时候还在上面疯狂答题的时间。
最大子列和是2025 MOOC杭州大学 数据结构课程问题集的第一道题
第三种方法是陈越老师在慕课里面讲的第三种方法。
言归正传，这次浙江大学的PTA数据结构的习题集第一题最大子列和，编译的过程之中就是报错：ignoring return value of ‘scanf’ declared with attribute ‘warn_unused_result’
就是说如果说键盘输入的数据可能会出现问题
第一点：输入数组长度的时候就是输入了数据。
第二点：以此输入每一个数组的元素
这两个地方报错，所以就是还是有很多的细节需要在下去注意的
修改之前的代码如下：
#include<stdio.h>
#include<stdlib.h>
#define MAXSIZE 100000 /* The Max size of the array length */
/**
* @param A is the one of the three number in the compare
* @param B is the one of the three number in the compare
* @param C is the one of the three number in the compare
*/
int Max3(int A, int B, int C)
{
/* return the biggest number from 3 digit */
return A > B ? A > C ? A : C : B > C ? B : C;
}
/**
* @param List is the array List
* @param left is the left index
* @param right is the right index
*/
int DivideAndConquer(int List[], int left, int right)
{
/* divide and conquer method to know List[left] to List[right] 's max subseq sum */
int MaxLeftSum = 0; /* The Left Max subseq sum */
int MaxRightSum = 0; /* The Right Max subseq sum */
int MaxLeftBorderSum = 0; /* The Left Max subseq sum on the border */
int MaxRightBorderSum = 0; /* The Right Max subseq sum on the border */
int LeftBorderSum = 0; /* The Left Border sum */
int RightBorderSum = 0; /* The Right Border sum */
int center = 0; /* The border */
int i = 0; /* The index */
if(left == right)
{ /* the condition to terminate the recursion : left a number */
if( List[left] > 0)
{
return List[left];
}
else
{
return 0;
}
}
/* The solution to divide */
center = (left + right) / 2; /* find the mid center point */
/* recure to calcuate the max subseq sum */
MaxLeftSum = DivideAndConquer(List, left, center);
MaxRightSum = DivideAndConquer(List, center+1, right);
/* cross the border to calcuate the max subseq sum */
MaxLeftBorderSum = 0;
LeftBorderSum = 0;
for(i = center; i >= left; i--)
{ /* from center to left scan */
LeftBorderSum += List[i];
if(LeftBorderSum > MaxLeftBorderSum)
{
MaxLeftBorderSum = LeftBorderSum;
}
} /* left side scan terminated */
MaxRightBorderSum = 0;
RightBorderSum = 0;
for(i = center + 1; i <= right; i++)
{ /* from center to right scan */
RightBorderSum += List[i];
if(RightBorderSum > MaxRightBorderSum)
{
MaxRightBorderSum = RightBorderSum;
}
} /* right side scan terminated */
/* Then we show the conquer result */
return Max3(MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum);
}
/**
* @param List is the array to be examine
* @param N is the length of the array List
*/
int MaxSubseqSum3(int List[], int N)
{
/* keep the same interface with 2 before algorithm */
return DivideAndConquer(List, 0, N - 1);
}
int main(int argc, char *argv[])
{
int length = 0;
scanf("%d",&length);
/* Just use malloc function to seek the enought memory space to fill the array */
int *array = (int*)malloc(length * sizeof(int));
if(array == NULL) // Memory malloc failed!
{
printf("Memory allocation failed!\n");
}
// fixed: The index is i, avoid to overflow
for(int i = 0; i < length; i++)
{
scanf("%d",&array[i]);
}
int result = MaxSubseqSum3(array, length);
printf("%d\n", result);
/* !!! IMPORTANT OPERATION, free the memory which allocate
* Please Just Do it *
* It is very important */
free(array);
return 0;
}
编译提示警告：
a.c: In function ‘main’:
a.c:93:5: warning: ignoring return value of ‘scanf’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
93 | scanf("%d",&length);
| ^~~~~~~~~~~~~~~~~~~
a.c:104:13: warning: ignoring return value of ‘scanf’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
104 | scanf("%d",&array[i]);
| ^~~~~~~~~~~~~~~~~~~~~
在网络上面就是查询很多的方法，最直接的方法就是检查每一次scanf("%d", &number)的结果是否为1
这是一个在下之前从来都不会想到的，但是从某种意义上面来说，这个也是在下已经忽略了很久的东西。
网络上面有非常多的为了解决而解决这个问题的方法。但是个人并不欣赏这些方法。
还有就是程序注释。很多的程序员不喜欢写注释，但是看别人的代码的时候又不喜欢别人不写注释，这个也是个人需要注意的。
还有就是将各个函数的参数都写清楚名称和作用，向别人描述清楚函数作用
C DocString编码风格和规范，doxygen也是个人之后需要学习的
列一下更改之后的代码：
#include<stdio.h>
#include<stdlib.h>
#define MAXSIZE 100000 /* The Max size of the array length */
/**
* @param A is the one of the three number in the compare
* @param B is the one of the three number in the compare
* @param C is the one of the three number in the compare
*/
int Max3(int A, int B, int C)
{
/* return the biggest number from 3 digit */
return A > B ? A > C ? A : C : B > C ? B : C;
}
/**
* @param List is the array List
* @param left is the left index
* @param right is the right index
*/
int DivideAndConquer(int List[], int left, int right)
{
/* divide and conquer method to know List[left] to List[right] 's max subseq sum */
int MaxLeftSum = 0; /* The Left Max subseq sum */
int MaxRightSum = 0; /* The Right Max subseq sum */
int MaxLeftBorderSum = 0; /* The Left Max subseq sum on the border */
int MaxRightBorderSum = 0; /* The Right Max subseq sum on the border */
int LeftBorderSum = 0; /* The Left Border sum */
int RightBorderSum = 0; /* The Right Border sum */
int center = 0; /* The border */
int i = 0; /* The index */
if(left == right)
{ /* the condition to terminate the recursion : left a number */
if( List[left] > 0)
{
return List[left];
}
else
{
return 0;
}
}
/* The solution to divide */
center = (left + right) / 2; /* find the mid center point */
/* recure to calcuate the max subseq sum */
MaxLeftSum = DivideAndConquer(List, left, center);
MaxRightSum = DivideAndConquer(List, center+1, right);
/* cross the border to calcuate the max subseq sum */
MaxLeftBorderSum = 0;
LeftBorderSum = 0;
for(i = center; i >= left; i--)
{ /* from center to left scan */
LeftBorderSum += List[i];
if(LeftBorderSum > MaxLeftBorderSum)
{
MaxLeftBorderSum = LeftBorderSum;
}
} /* left side scan terminated */
MaxRightBorderSum = 0;
RightBorderSum = 0;
for(i = center + 1; i <= right; i++)
{ /* from center to right scan */
RightBorderSum += List[i];
if(RightBorderSum > MaxRightBorderSum)
{
MaxRightBorderSum = RightBorderSum;
}
} /* right side scan terminated */
/* Then we show the conquer result */
return Max3(MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum);
}
/**
* @param List is the array to be examine
* @param N is the length of the array List
*/
int MaxSubseqSum3(int List[], int N)
{
/* keep the same interface with 2 before algorithm */
return DivideAndConquer(List, 0, N - 1);
}
int main(int argc, char *argv[])
{
int length = 0;
if(scanf("%d",&length) == 1)
{
}
else
{
printf("Failed to read integer.\n");
}
/* Just use malloc function to seek the enought memory space to fill the array */
int *array = (int*)malloc(length * sizeof(int));
if(array == NULL) // Memory malloc failed!
{
printf("Memory allocation failed!\n");
}
// fixed: The index is i, avoid to overflow
for(int i = 0; i < length; i++)
{
if(scanf("%d",&array[i]) == 1)
{
}
else
{
printf("Failed to read integer.\n");
}
}
int result = MaxSubseqSum3(array, length);
printf("%d\n", result);
/* !!! IMPORTANT OPERATION, free the memory which allocate
* Please Just Do it *
* It is very important */
free(array);
return 0;
}
还有就是最后一点的Free，老师在上课的时候反复强调过就是malloc和calloc的函数都是需要释放内存的，但是个人在查询了很多资料的时候还是发现有的网站上面并没有写free函数，就是最好还是写一下，能不能执行是另外一方面，但是规范是在那里的。
谢谢老师和同学们不厌其烦的帮助。