119. Pascal's Triangle II

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119. Pascal's Triangle II

题目

Given an index k, return the kth row of the Pascal's triangle.

For example, given k = 3, //其实为第四行
Return [1,3,3,1].

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

     [1],
    [1,1],
   [1,2,1],
  [1,3,3,1],
 [1,4,6,4,1]

解析

• 注意使用逆序累计
• 第二种方法可以利用数学的递推关系来完成：If anyone has ever learnt the mathematics equations related to the pascal triangle, they would know the following:The nth row of pascal triangle will have the following format: 1 a(1) a(2) ... a(n) here we have a(1) = n; a(k+1) = a(k) * (n-k)/(k+1).

class Solution {
public:
       //  从后往前迭代
	vector<int> getRow1(int rowIndex) {

		vector<int> dp(rowIndex + 1, 1);

		for (int i = 2; i<rowIndex + 1; i++) {
			for (int j = i - 1; j>0; j--)
				dp[j] = dp[j] + dp[j - 1];

		}
		return dp;
	}

    vector<int> getRow(int rowIndex) {
		//A[i]=A[i-1]+A[i]      0<i<n-1
		vector<int> A;
		if (rowIndex < 0)  
			return A;
		A.resize(rowIndex + 1, 0);
		A[0] = 1; //第一行的数
		for (int k = 1; k<=rowIndex; k++){
			for (int j = k; j>0; j--){
				if (j == k)   
					A[j] = 1; //每行结尾的数
				else{
					A[j] = A[j] + A[j - 1];
				}
			}
		}
		return A;
	}
};

题目来源

• 119. Pascal's Triangle II
• leetcode: Pascal's Triangle II