LeetCode(119)Pascal's Triangle II
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题目如下:
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?
分析如下:
注意题目要求是O(K)的空间,所以不能向上一题一样打印出一个二维数组了,可以维护一个一维数组,并且不断地修改它,使得它的成为第k行的Pascal's Triangle.
我的代码:
class Solution {public: vector<int> getRow(int rowIndex) { vector<int> res; if(rowIndex<0) return res; res.push_back(1); for(int i=1;i<rowIndex+1;i++){ int pre=0; int pre_size=(int)res.size(); for(int j=0;j<pre_size;j++){ int cur=res[j]; res[j]=pre+res[j]; pre=cur; } res.push_back(1); } return res; }};
update: 简单的一维动态规划
03-08-2015
// 2ms class Solution {public: vector<int> getRow(int rowIndex) { //看清楚题,k从0开始计数,不是从1开始计数。 vector<int> result; if (rowIndex < 0) return result; result.push_back(1); for (int i = 1; i <= rowIndex; ++i) { result.push_back(result.back()); for (int j = i - 1; j > 0; --j) { result[j] += result[j - 1]; } } return result; }};
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