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+ /*
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+ Given an index k, return the kth row of the Pascal's triangle.
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+
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+ For example, given k = 3,
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+ Return [1,3,3,1].
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+
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+ Note:
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+ Could you optimize your algorithm to use only O(k) extra space?*/
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+
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+ public class Solution {
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+ public ArrayList <Integer > getRow (int rowIndex ) {
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+ // Start typing your Java solution below
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+ // DO NOT write main() function
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+ ArrayList <ArrayList <Integer >> triangle = new ArrayList <ArrayList <Integer >>();
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+ for (int i = 0 ; i <= rowIndex ; i ++){
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+ ArrayList <Integer > tmp = new ArrayList <Integer >();
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+ tmp .add (1 );
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+ if (i > 0 ){
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+ for (int j = 0 ; j < triangle .get (i - 1 ).size () - 1 ; j ++){
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+ tmp .add (triangle .get (i - 1 ).get (j ) + triangle .get (i - 1 ).get (j + 1 ));
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+ }
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+ tmp .add (1 );
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+ }
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+ triangle .add (tmp );
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+ }
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+ return triangle .get (rowIndex );
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+ }
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+ }
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+
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+
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+ //********************O(k) Solution
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+ /*
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+ 如果没有这个O(k)空间的限制,那么可以一行一行迭代生成。如果要直接生成第i行,假设生成k=3,可以这样考虑这样的一个过程:
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+ 1 0 0 0 k = 0
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+ 1 1 0 0 k = 1
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+ 1 1 1 0
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+ 1 2 1 0 k = 2
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+ 1 2 1 1
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+ 1 2 3 1
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+ 1 3 3 1 k = 3
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+ 上述过程实际上就是一个in-place的迭代过程。每当生成下一行的时候,首先数组相应位置1,然后从右向左计算每一个系数。
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+ http://blog.csdn.net/abcbc/article/details/8982651
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+ */
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+
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+ public ArrayList <Integer > getRow (int rowIndex ) {
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+ // Start typing your Java solution below
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+ // DO NOT write main() function
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+ ArrayList <Integer > result = new ArrayList <Integer >(rowIndex + 1 );
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+ for (int i = 0 ; i <= rowIndex ; i ++) {
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+ result .add (0 );
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+ }
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+ result .set (0 , 1 );
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+ for (int i = 1 ; i <= rowIndex ; i ++) {
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+ result .set (i , 1 );
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+ for (int j = i - 1 ; j > 0 ; j --) {
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+ result .set (j , result .get (j ) + result .get (j - 1 ));
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+ }
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+ }
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+ return result ;
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+ }
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