import java.util.Arrays;
/**
*
* Source : https://oj.leetcode.com/problems/pascals-triangle-ii/
*
*
* 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?
*
*/
public class PascalTriangle2 {
/**
* 获取杨辉三角的第n行
* 占用常数空间,使用长度为n的数组,依次计算 ...i-2,i-2,i 行的数据,可以根据俄第i-1行的结果来计算i行,
* 计算每行数据的时候,如果从前向后填入的话,会把后面的覆盖就不能继续计算,那么就从后向前
*
* @param n
* @return
*/
public int[] getRow (int n) {
int[] row = new int[n];
row[0] = 1;
if (n == 1) {
return row;
}
for (int i = 2; i <= n; i++) {
for (int j = i; j > 1; j--) {
row[j-1] = row[j-1] + row[j-2];
}
}
return row;
}
public static void main(String[] args) {
PascalTriangle2 pascalTriangle2 = new PascalTriangle2();
System.out.println(Arrays.toString(pascalTriangle2.getRow(5)));
}
}