LeetCode: Pascal's Triangle II
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思路:因为只能使用O(k)的空间,所以预先分配 k 个元素容量的数组,然后再此数组上进行元素更新,根据公式ret(i,j) = ret(i-1,j-1) + ret(i-1, j)。
code:
class Solution {public: vector<int> getRow(int rowIndex) { vector<int> ret(rowIndex+1,1); for(int i=0;i<= rowIndex;i++) for(int j=i-1;j>0;j--) ret[j] = ret[j] + ret[j-1]; return ret; }};
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