leetcode 118. Pascal's Triangle 杨辉三角形
2017-09-14 13:20
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Given numRows, generate the first numRows of Pascal’s triangle.
For example, given numRows = 5,
Return
[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]
这个就是中国最伟大的杨辉三角形的问题。
代码如下:
下面是C++的做法,这个是中国最经典的杨辉三角形的问题
代码如下:
For example, given numRows = 5,
Return
[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]
这个就是中国最伟大的杨辉三角形的问题。
代码如下:
import java.util.ArrayList; import java.util.List; /* * 中国最伟大的杨辉三角形 * */ public class Solution { public List<List<Integer>> generate(int numRows) { List<List<Integer>> res=new ArrayList<>(); if(numRows==0) return res; else if(numRows==1) { List<Integer> one=new ArrayList<>(); one.add(1); res.add(one); return res; }else { List<Integer> one=new ArrayList<>(); one.add(1); one.add(0); res.add(one); for(int i=1;i<numRows;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; } } }
下面是C++的做法,这个是中国最经典的杨辉三角形的问题
代码如下:
#include <iostream> #include <string> #include <vector> #include <queue> using namespace std; class Solution { public: vector<vector<int>> generate(int n) { vector<vector<int>> res; if (n == 0) return res; else if (n == 1) { vector<int> one{ 1 }; res.push_back(one); return res; } else { vector<int> one{ 1 ,0}; res.push_back(one); for (int i = 1; i < n; 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; } } };
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