[LeetCode][Java] Pascal's Triangle II
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题目:
Given an index k, return the kth row of the Pascal's triangle.
For example, given k = 3,
Return [1,3,3,1]
.
题意:
给定序号 k,返回杨辉三角中的第k行。
比如,给定k = 3,
返回[1,3,3,1]
.
算法分析:
与题目《Pascal's Triangle》算法类似,每个元素等于上一行的对应两元素的和值,这里只是不把每一行都保存,只需推导到所求行即可。
AC代码:
<span style="font-family:Microsoft YaHei;font-size:12px;">public class Solution{public static List<Integer> getRow(int rowIndex) { List<Integer> a=new ArrayList<Integer>(); for (int i=0;i<=rowIndex;i++) { Integer[] ta =new Integer[a.size()]; a.toArray(ta); for(int j=0;j<i+1;j++) { if(j==i) a.add(j,1); else if (j==0) a.set(j,1); else { int temp=ta[j-1]+ta[j]; a.set(j,temp); } } } return a; }}</span>
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