LeetCode 楊輝三角 題目解答
/**
* Return an array of arrays.
* The sizes of the arrays are returned as *columnSizes array.
* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
*/
int** generate(int numRows, int** columnSizes) {
printf("row : %d\n",numRows);
int row = numRows;
int** p = NULL;
if(numRows == 0)
{
*columnSizes = NULL;
return p;
}
p = (int**)malloc(sizeof(int*)*row);
*columnSizes = (int*)malloc(sizeof(int)*row);
for(int i = 0;i<row;i++)
{
int j = i+1;
p[i] = (int*)malloc(sizeof(int)*j);
// columnSizes[i]=(int*)malloc(sizeof(int));
(*columnSizes)[i] = j;
if(i == 0)//first row
{
p[i][0] = 1;
printf("%d \n",p[i][0]);
}
else
{
// printf("colsize :%d\n",j);
for(int k = 0;k<j;k++)
{
if(k == 0||k == j-1)
{
p[i][k] = 1;
printf("%d \t",p[i][k]);
if(k == j-1)
{
printf("\n");
}
}
else
{
p[i][k] = p[i-1][k-1]+p[i-1][k];
printf("%d \t",p[i][k]);
}
}
}
}
return p;
}
思考:
numRows:輸入的行數
columnSize:二維陣列用於存放 每一行元素的個數(真是坑,如果這個值沒做好,那麼提交是讀不出每行的元素的)
*columnSizes = (int*)malloc(sizeof(int)*row);
(*columnSizes)[i] = j; 記錄每一行的 元素個數=行數+1
函式裡:
1.注意極限值:例如 輸入 row = 0;則返回NULL指標
2.邊界元素均為 1