Pascal Triangle

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Given an index k, return the k^th 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?

10/31

这道题挺简单的，还是犯了两个小错，见err1 和 err2

    public List<Integer> getRow(int n) {        // err1: ask if n could be 0?        List<Integer> row = new ArrayList<Integer>();        row.add(1);        for(int i=0; i<n; i++){            List<Integer> copy = new ArrayList<Integer>(row);            row = new ArrayList<Integer>();            row.add(1);            int len = copy.size();            for(int j=0; j<len-1; j++){ //err2: use a different index other than j                row.add(copy.get(j) + copy.get(j+1));            }            row.add(1);        }        // directly return [1] on n=0                return row;    }}

时间复杂度：

一共有n个循环，在第i个循环生成第i行，大约要做i个加法，所以复杂度大概是O(n^2)

空间复杂度：

在每个循环中需要先储存上一个循环中得到的结果，加上每次都需要记录目前的结果，所以复杂度是O(n)\

Follow up: Further improvement:

Without extra space to store the previous row, we can scan each row backwards

public class Solution {    public List<Integer> getRow(int n) {        // err1: ask if n could be 0?        List<Integer> row = new ArrayList<Integer>();        row.add(1);        for(int i=0; i<n; i++){            int len = row.size();            for(int j=len-1; j>0; j--){                row.set(j, row.get(j) + row.get(j-1));            }                        row.add(1); // err1 : only need to append to the end; the first element is always 1: never reset        }        // directly return [1] on n=0                return row;    }}

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