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Leetcode-Pascal's Triangle II

时间:2014-11-08 10:25:28      阅读:218      评论:0      收藏:0      [点我收藏+]

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?

 

Solution:

public class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> preRow = new ArrayList<Integer>();
        List<Integer> curRow = new ArrayList<Integer>();
        
        curRow.add(1);
        int len;
        for (int i=1;i<=rowIndex;i++){
            preRow = curRow;
            curRow = new ArrayList<Integer>();
            len = i+1;
            //j==0
            curRow.add(1);
            //j==1 to (len-2)
            for (int j=1;j<len-1;j++)
                curRow.add(preRow.get(j-1)+preRow.get(j));
            //j==len-1
            curRow.add(1);
        }
        
        return curRow;
    }
}

递推问题。只需存当前行和上一行就行。

Leetcode-Pascal's Triangle II

原文:http://www.cnblogs.com/lishiblog/p/4082728.html

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