leetcode 118 Pascal's Triangle || 119 Pascal's Triangle II

本次题解包括

• 118. Pascal's Triangle
• 119. Pascal's Triangle II

118. Pascal's Triangle

Given numRows, generate the first numRows of Pascal's triangle.

For example, given numRows = 5, Return

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

题目地址：leetcode Pascal's Triangle

题目大意：给定n返回前n行杨辉三角

思路：看代码吧

C++

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> ans;
        if(numRows < 1) return ans;
        ans.push_back(vector<int>{1});
        for(int i = 1; i < numRows; ++i){
            vector<int> cur(i + 1, 0);
            for(int j = 0; j <= i; ++j)
                cur[j] = (j != i? ans[i - 1][j]:0) + (j > 0? ans[i - 1][j - 1]: 0);
            ans.push_back(cur);
        }
        return ans;
    }
};

Python

class Solution(object):
    def generate(self, numRows):
        """
        :type numRows: int
        :rtype: List[List[int]]
        """
        ans=[[1], [1,1]]
        for i in range(1, numRows + 1):
            temp = [1]+ [ans[-1][t] + ans[-1][t + 1] for t in range(len(ans[-1]) - 1)] + [1]
            ans.append(temp)
        return ans[:numRows]

 

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119. Pascal's Triangle II

Given an index k, return the _k_th 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?

题目地址：leetcode Pascal's Triangle II

题目大意：给定n返回第n行杨辉三角(下标从0开始)

思路：看代码

C++

class Solution {
public:
    vector<int> getRow(int k) {
        vector<int> cur(k + 1, 0);
        cur[0] = 1;
        for(int i = 1; i <= k; ++i){
            vector<int> next(i + 1, 1);
            for(int j = 1; j < i; ++j)
                next[j] = cur[j - 1] +  cur[j]; 
            cur = next;
        }
        return cur;
    }
};

Python

class Solution(object):
    def getRow(self, rowIndex):
        """
        :type rowIndex: int
        :rtype: List[int]
        """
        ans = [[1], [1, 1]]
        if rowIndex < 2: return ans[rowIndex]
        pre = ans[-1]
        for i in range(1, rowIndex):
            temp = [1] + [pre[t] + pre[t + 1] for t in range(len(pre) - 1)] + [1]
            pre = temp
        return pre

本文是leetcode如下的题解

• 118. Pascal's Triangle
• 119. Pascal's Triangle II

更多题解可以查看： https://www.hrwhisper.me/leetcode-algorithm-solution/