翻译
给定一个索引K,返回帕斯卡三角形的第K行。
例如,给定K=3,
返回[1,3,3,1]。
注释:
你可以改进你的算法只用O(k)的额外空间吗?
原文
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?
分析
这一题呢实际上是承接上一题的,我也正好刚写完。
LeetCode 118 Pascal’s Triangle(帕斯卡三角形)(vector)
上一题是返回完整的帕斯卡三角形:
class Solution {
public:
vector<vector<int>> generate(int numRows) {
vector<vector<int>> pascal;
if (numRows < 1) return pascal;
vector<int> root;
root.push_back(1);
pascal.push_back(root);
if (numRows == 1) return pascal;
root.push_back(1);
pascal.push_back(root);
if (numRows == 2) return pascal;
if (numRows > 2) {
for (int i = 2; i < numRows; ++i) {
vector<int> temp;
temp.push_back(1);
for (int j = 1; j < pascal[i - 1].size(); ++j) {
temp.push_back(pascal[i - 1][j - 1] + pascal[i - 1][j]);
}
temp.push_back(1);
pascal.push_back(temp);
}
return pascal;
}
}
};
所以我就偷个懒,既然是索引K,那就返回它好了,不过效率就……
class Solution {
public:
vector<int> getRow(int rowIndex) {
rowIndex += 1;
vector<vector<int>> pascal;
if (rowIndex < 1) return pascal[0];
vector<int> root;
root.push_back(1);
pascal.push_back(root);
if (rowIndex == 1) return pascal[0];
root.push_back(1);
pascal.push_back(root);
if (rowIndex == 2) return pascal[1];
if (rowIndex > 2) {
for (int i = 2; i < rowIndex; ++i) {
vector<int> temp;
temp.push_back(1);
for (int j = 1; j < pascal[i - 1].size(); ++j) {
temp.push_back(pascal[i - 1][j - 1] + pascal[i - 1][j]);
}
temp.push_back(1);
pascal.push_back(temp);
}
return pascal[rowIndex - 1];
}
}
};
更高效的方法应该是有某种公式了,看看别人写的就知道了……
vector<int> getRow(int rowIndex) {
vector<int> r;
r.resize(rowIndex + 1);
r[0] = r[rowIndex] = 1;
for (auto i = 1; i < (r.size() + 1) / 2; ++i) {
r[i] = r[rowIndex - i] = (unsigned long)r[i - 1] * (unsigned long)(rowIndex - i + 1) / i;
}
return r;
}
果然,数学的威力再次显现了。让我用Markdown语法写出这个公式……
ri=ri−1∗(index−i+1)/i
Java, updated at 2016/8/28
public class Solution {
public List<Integer> getRow(int rowIndex) {
List<Integer> list = new ArrayList<Integer>();
if (rowIndex < 0) return list;
for (int i = 0; i <= rowIndex; i++) {
list.add(0, 1);
for (int j = 1; j < list.size() - 1; j++) {
list.set(j, list.get(j) + list.get(j + 1));
}
}
return list;
}
}