leetcode之Pascal's Triangle II
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Pacsal’s Triangle也称杨辉三角,其每一行的值都是其肩上两个数的和,leetcode中求杨辉三角的算法比较简单,在此不再表述,下面对Pasca‘s Triangle II进行分析,其原题如下:
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?
http://blog.csdn.net/fm0517/article/details/4254099
代码实现如下:
class Solution {public: vector<int> getRow(int rowIndex) { vector<int>v(rowIndex+1,1);if(rowIndex == 0 || rowIndex == 1)return v;for(int i = 2; i <= rowIndex; i++ ){for(int j = i - 1; j > 0; j--){ v[j] = v[j] + v[j-1];}}return v; }};今天A的leetcode上的几道题都是空间限制类型的,在空间受限的情况下,要尽最大可能的发挥已有空间的使用效率,并注意计算顺序,如今天上午那道腾讯2012年笔试题中的计算乘积的题目是通过前后两种遍历实现的,而矩阵置0是将第一行和第一列作为标志使用,而当前这道题则是通过倒序计算来节省空间。
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