LeetCode:Pascal's Triangle II

Given an index k, return the k^th 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?

解题思路:

    由于计算杨辉三角时，只用到相邻的两行的数据，所以我们可以反向计算，就能以O(k)

的时间复杂度解决问题了.

解题代码:

class Solution {
public:
    vector<int> getRow(int rowIndex) 
    {
        vector<int> res;
        for (int i = 0; i <= rowIndex; ++i)
        {
            for (int j = i - 1; j > 0; --j)
                res[j] = res[j] + res[j-1];
            res.push_back(1);
        }
        return res;
    }
};

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LeetCode:Pascal's Triangle II

原文：http://blog.csdn.net/dream_you_to_life/article/details/32715163