LEETCODE--Pascal's Triangle II

最新推荐文章于 2024-05-25 16:17:00 发布

prettysky123

最新推荐文章于 2024-05-25 16:17:00 发布

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本文介绍两种高效算法来获取帕斯卡三角形的第K行元素，重点在于如何仅使用O(k)额外空间实现该功能。通过具体代码示例展示了不同方法的实现细节。

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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?

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方法一：
从后往前进行加法赋值；

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> A; 
        A.push_back(1);
        if(rowIndex == 0)
            return A;
        A.push_back(1);
        if(rowIndex == 1)
            return A;
        for(int x = 2; x <= rowIndex; x++){
           A.push_back(0);
        }
        for(int i = 2; i <= rowIndex; i++){
            A[i] = 1;
            for(int j = i-1; j > 0; j--){
                A[j] = A[j] + A[j-1];    
            }
            A[0] = 1;
        }
        return A;
    }
};

方法二：

class Solution {
public:
    vector<int> getRow(int rowIndex) {

        vector<vector<int>> finsh;
        vector<int> last;
        vector<int> add;
        finsh.push_back(add);
        finsh[0].push_back(1);
        rowIndex += 1;
        if(rowIndex == 1)
            return finsh[0];
        else{
              for(int i = 1; i < rowIndex; i++){
                    if(i == finsh.size())
                    {
                        vector<int> add;
                        finsh.push_back(add);
                    } 
                    finsh[i].push_back(1);
                    for(int j = 1; j <= i; j++){
                        if(finsh[i].size() == i)
                            finsh[i].push_back(1);
                        else{
                            int x = finsh[i-1][j-1];
                            int y = finsh[i-1][j];
                            int elem = x + y;
                            finsh[i].push_back(elem);
                        }
                    }
               }
               return finsh[rowIndex-1];
         }

    }
};