119. 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?
首先思考时间复杂度与空间复杂度,这样更能理解别人的代码比你优秀在哪。
题目要求空间复杂度为O(k),然后画图想想遍历的顺序,从第0行至第k行,每行从0至k
则算法的时间复杂度O(k^2)
想了一会写下了如下代码:
class Solution {public: vector<int> getRow(int rowIndex) { if(rowIndex<0) return vector<int>(); vector<int> dp(rowIndex+1); dp[0]=1; for(int row=1;row<rowIndex+1;row++ for(int i=1;i<=row;i++) dp[i]=dp[i-1]+dp[i]; return dp; }};结果报错了,看了Discuss才恍然大悟,可以这么写
class Solution {public: vector<int> getRow(int rowIndex) { if(rowIndex<0) return vector<int>(); vector<int> dp(rowIndex+1); dp[0]=1; for(int row=1;row<rowIndex+1;row++ for(int i=row;i>=1;i--) dp[i]=dp[i-1]+dp[i]; return dp; }};
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