leetcode 119. Pascal's Triangle II 杨辉三角形2
2017-09-14 13:22
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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].
就是中国最伟大的杨辉三角形。
代码如下:
下面是C++的做法,和上面的一道题一模一样,就是做一个迭代计算即可完成
代码如下:
For example, given k = 3,
Return [1,3,3,1].
就是中国最伟大的杨辉三角形。
代码如下:
import java.util.ArrayList; import java.util.List; public class Solution { public List<Integer> getRow(int rowIndex) { List<List<Integer>> res=new ArrayList<>(); if(rowIndex==0) { List<Integer> one=new ArrayList<>(); one.add(1); res.add(one); return res.get(0); }else { List<Integer> one=new ArrayList<>(); one.add(1); one.add(0); res.add(one); for(int i=1;i<rowIndex+1;i++) { List<Integer> tt=res.get(i-1); List<Integer> rr=new ArrayList<>(); rr.add(tt.get(0)); for(int j=1;j<tt.size();j++) rr.add(tt.get(j) + tt.get(j-1)); rr.add(0); res.add(rr); } for(int i=0;i<res.size();i++) res.get(i).remove(res.get(i).size()-1); return res.get(rowIndex); } } }
下面是C++的做法,和上面的一道题一模一样,就是做一个迭代计算即可完成
代码如下:
#include <iostream> #include <vector> using namespace std; class Solution { public: vector<int> getRow(int rowIndex) { if (rowIndex == 0) return vector<int>(1,1); else { vector<vector<int>> res; vector<int> one{ 1 ,0 }; res.push_back(one); for (int i = 1; i < rowIndex+1; i++) { vector<int> one{ res[i - 1][0] }; for (int j = 1; j < res[i - 1].size(); j++) one.push_back(res[i - 1][j] + res[i - 1][j - 1]); one.push_back(0); res.push_back(one); } for (int i = 0; i < res.size(); i++) res[i].erase(res[i].end() - 1); return res[rowIndex]; } } };
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