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Pascal‘s Triangle II
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?
解法一:利用Pascal‘s Triangle的结果,选取最后一行返回即可
class Solution { public: vector<int> getRow(int rowIndex) { vector<vector<int> > result = generate(rowIndex+1); return result[rowIndex]; } vector<vector<int> > generate(int numRows) { vector<vector<int> > result; if(numRows == 0) { //nothing } else if(numRows == 1) { //第0层 vector<int> cur; cur.push_back(1); result.push_back(cur); } else { //第0层 vector<int> cur; cur.push_back(1); result.push_back(cur); for(int n = 1; n < numRows; n ++) {//用<n choose m>的通项公式来做肯定溢出,所以只能用加法做 vector<int> cur; //头 cur.push_back(1); //中间元素是上层两元素相加 for(int i = 0, j = 1; j < n; i++, j++) { vector<int> pre = result[result.size()-1]; cur.push_back(pre[i]+pre[j]); } //尾 cur.push_back(1); result.push_back(cur); } } return result; } };
【LeetCode】Pascal's Triangle II
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原文地址:http://www.cnblogs.com/ganganloveu/p/4128953.html