【49】119. Pascal's Triangle II

119. Pascal's Triangle II

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• Total Accepted: 102894
• Total Submissions: 291602
• Difficulty: Easy
• Contributors: Admin

 

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?

class Solution {
public:
//杨辉三角从0层开始往下，第n层的第k个是Cnk（k在上面）
//本地调试输出前十行，没啥问题，拿到OJ上测试，程序在第18行跪了，中间有个系数不正确。那么问题出在哪了呢，仔细找找，原来出在计算组合数那里，由于算组合数时需要算连乘，而整形数int的数值范围只有-32768到32768之间，那么一旦n值过大，连乘肯定无法计算。
/*
   vector<int> getRow(int rowIndex) {
        vector<int> out;
        if (rowIndex < 0) return out;
        
        for (int i = 0; i <= rowIndex; ++i) {
            if ( i == 0 || i == rowIndex)
                out.push_back(1);
            else
                out.push_back (computeCnk(rowIndex, i));
        }
        return out;
    }
    
    int computeCnk(int n, int k) {
        if (k > n) return 0;
        else if (k > n/2) k = n - k;
        int numerator = 1, denomator = 1;
        for (int i = 0; i < k; ++i) {
            numerator *= n - i;
            denomator *= k - i;
        }
        if (denomator != 0) return numerator/denomator;
        else return 0;
    }*/
    
    vector<int> getRow(int rowIndex) {
        vector<int> out(rowIndex + 1, 0);
        //if (rowIndex < 0) return out;
        
        //out.assign(rowIndex + 1, 0);
        for (int i = 0; i <= rowIndex; ++i) {
            if ( i == 0) {
                out[0] = 1;
                continue;
            }
            for (int j = rowIndex; j >= 1; --j) {
                out[j] = out[j] + out[j-1];
            }
        }
        return out;
    }
   
};