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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?
1 public class Solution { 2 public List<Integer> getRow(int rowIndex) { 3 Integer[] num = new Integer[rowIndex+1]; 4 num[0] = 1; 5 for(int i = 1; i < num.length; i++){ 6 num[i] = (int)((long)num[i-1]*(rowIndex-i+1)/i);//不加强制转换,乘积大于Integer.MAX_VALUE有溢出 7 } 8 return Arrays.asList(num); 9 } 10 }
1 public class Solution { 2 public List<Integer> getRow(int rowIndex) { 3 List<Integer> list = new ArrayList<Integer>(); 4 list.add(1); 5 for(int i = 1; i < rowIndex + 1; i++){ 6 list.add((int)((long)list.get(i-1)*(rowIndex-i+1)/i)); 7 } 8 return list; 9 } 10 }
根据数学公式:
[C(k,0), C(k,1), ..., C(k, k-1), C(k, k)]
C[k,i] = C[k,i-1]*(k-i+1)/i
119. Pascal's Triangle II java solutions
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原文地址:http://www.cnblogs.com/guoguolan/p/5646375.html