LEET code -- Pascal's Triangle II

Given an index k, return the k^th row of the Pascal's triangle.
For example, given k = 3,
Return [1,3,3,1].

Note:
Could you optimize your algorithm to use only O(k) extra space?

我本來還想找規律
但是這根本沒辦法找規律
因為這一定要用之前算好的累積

所以可以只用一個陣列 但是存的時候是倒過來
因為夏禕牌都比上一排多一個   所以可以不蓋過需要的element

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */ int* getRow(int rowIndex, int* returnSize) { //store the element in the tail, to save the space if(rowIndex<0)return NULL; *returnSize=rowIndex+1; int *ans=malloc(sizeof(int)*(rowIndex+1)); for(int i=0;i<=rowIndex;i++){ for(int j=i;j>=0;j--){ if(j==0 || j==i) ans[j]=1; else{ ans[j]=ans[j]+ans[j-1]; } } } return ans; }