leetcode 119. Pascal's Triangle II

Given a non-negative index k where k ≤ 33, return the k^th index row of the Pascal's triangle.

Note that the row index starts from 0.

In Pascal's triangle, each number is the sum of the two numbers directly above it.

Example:

Input: 3
Output: [1,3,3,1]

Follow up:

Could you optimize your algorithm to use only O(k) extra space?

class Solution:
    def getRow(self, rowIndex):
        """
        :type rowIndex: int
        :rtype: List[int]
        """
        start = ans = [1]
        for i in xrange(0, rowIndex):            
            ans = start + [1]
            for j in xrange(1, len(ans)-1):
                ans[j] = start[j]+start[j-1]
            start = ans
        return ans

还可以少一个临时变量，从后向前计算相加：

class Solution:
    def getRow(self, rowIndex):
        """
        :type rowIndex: int
        :rtype: List[int]
        """
        ans = [1]
        for i in xrange(0, rowIndex):            
            ans = ans + [1]
            for j in xrange(len(ans)-2, 0, -1):
                ans[j] = ans[j]+ans[j-1]            
        return ans

也有使用dummy变量的做法：

class Solution(object):
    def getRow(self, rowIndex):
        """
        :type rowIndex: int
        :rtype: List[int]
        """
        row = [1]
        for _ in range(rowIndex):
            row = [x + y for x, y in zip([0]+row, row+[0])]
        return row

另外就是数学解法，没有懂，TODO：

class Solution {
public:
    vector<int> getRow(int k) {
        vector<int> ans(k+1,1);
        for(int i=1;i<=k/2;++i){          
           ans[k-i]= ans[i]=long(ans[i-1])*(k-i+1)/i;           
        }        
        return ans;
    }
};