LeetCode Pascal's Triangle II
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Given an index k, return the kth 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?
第0层为1,第1层为1 1,第2层为1 2 1,第3层为1 3 3 1,从第2层到第3层,a[0]=1不变,a[1]=原a[0]+a[1],a[2]=原a[1]+a[2],a[3]为添加进去的1,于是,可以用动态规划的思想,从第i层到第i+1层,a[0]=1不变,a[1]=原a[0]+a[1]...a[k]=原a[k-1]+a[k],最后在数组尾添加元素1。
class Solution {public: vector<int> getRow(int rowIndex) { vector<int> res; res.push_back(1); int cur,pre; for(int i=1;i<=rowIndex;i++){ pre=res[0]; for(int j=1;j<res.size();j++){ cur=res[j]; res[j]+=pre; pre=cur; } res.push_back(1); } return res; }};
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