[BackTracking]119. Pascal's Triangle II

• 分类：BackTracking
• 时间复杂度: O(n^2)
• 空间复杂度: O(n^2) or O(k)

119. Pascal's Triangle II

Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.

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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?

代码：

原来想的代码：

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        
        if rowIndex==0:
            return [1]
        prev=self.getRow(rowIndex-1)
        current=[1 for i in range(rowIndex+1)]
        for i in range(1,rowIndex):
            current[i]=prev[i-1]+prev[i]
        return current

迭代DP，自己写的O(k)代码：

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        
        if rowIndex==0:
            return [1]
        dp={0:[1 for i in range(rowIndex+1)],1:[1 for i in range(rowIndex+1)]}
        for j in range(1,rowIndex+1):
            for i in range(1,j):
                dp[j%2][i]=dp[(j-1)%2][i-1]+dp[(j-1)%2][i]
        return dp[(rowIndex)%2]

迭代，Youtube上小梦想家的代码：

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        
        res=[1]+[0]*rowIndex
        for i in range(1,rowIndex+1):
            for j in range(i,0,-1):
                res[j]=res[j]+res[j-1]
        
        return res

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讨论：

1.这道题O(k)的方法相当于一个DP的解法
2.小梦想家的方法真的很简洁很舒服orz，我佛