Pascal's Triangle I，II

题目来自于Leetcode
https://leetcode.com/problems/pascals-triangle/
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]
]

class Solution {
public:
vector<vector<int>> generate(int numRows) {
vector<vector<int> >res;
for(int i=0;i<numRows;i++)
{
vector<int>vec(i+1,1);
if(i>1)
for(int j=1;j<i;j++)
vec[j]=res[i-1][j-1]+res[i-1][j];
res.push_back(vec);
vector<int>().swap(vec);
}
return res;
}
};

Pascal's Triangle II

Total Accepted: 42320 Total
Submissions: 143760

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?

此处有内存要求尽管採用第一种方法能够ac可是明显不符合要求

class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<vector<int> >res;
for(int i=0;i<rowIndex+1;i++)
{
vector<int>vec(i+1,1);
if(i>1)
for(int j=1;j<i;j++)
vec[j]=res[i-1][j-1]+res[i-1][j];
res.push_back(vec);
vector<int>().swap(vec);
}
return res[rowIndex];

}
};

我们必须又一次设计算法。

第一想到的就是pascal三角形的系数会等于N行i列的值等于

（ r

n ）

可是

class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int>res(rowIndex+1,1);
if(rowIndex<2)
return res;
long long nth=1;
for(int i=1;i<rowIndex+1;i++)
nth*=i;
long long rth=1,n_rth=nth;
for(int i=1;i<rowIndex;i++)
{

n_rth/=(rowIndex-i+1);
res[i]=nth/rth/n_rth;
rth*=(i+1);
}
return res;
}
};

用来存储二项式系数的值非常easy在rowIndex=24时候就报错了

最后一种也就是正确的方法是利用分配的空间来计算的详细给出了k=5的详细描写叙述

class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int>res(rowIndex+1,1);
if(rowIndex<2)
return res;
int t1,t2;
for(int i=2;i<=rowIndex;i++)
{
t1=res[0];
t2=res[1];
for(int j=1;j<i+1;j++)
{
res[j]=t1+t2;
t1=t2;
t2=res[j+1];
}
res[i]=1;
}
return res;
}
};

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