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?
10/31
這道題挺簡單的, 還是犯了兩個小錯,見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;
}
}
時間複雜度:
一共有n個循環, 在第i個循環生成第i行, 大約要做i個加法, 所以複雜度大概是O(n^2)
空間複雜度:
在每個循環中需要先儲存上一個循環中得到的結果, 加上每次都需要記錄目前的結果, 所以複雜度是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;
}
}