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118. Pascal's Triangle

118. Pascal's Triangle

难度:Easy

内容

题目链接:https://leetcode.com/problems/pascals-triangle

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

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

Example:

Input: 5
Output:
[
     [1],
    [1,1],
   [1,2,1],
  [1,3,3,1],
 [1,4,6,4,1]
]

思路

该题比较简单,杨辉三角形,从第三行开始,除了首尾,中间元素的值可通过如下公式计算:cur[i] = prev[i] + prev[i-1]

计算当前行实现要点如下:

  • 拷贝上一行,并在末尾添加元素1
  • 倒着开始计算,除了首尾,cur[i] += cur[i-1]

代码

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> res;
        for (auto i = 0; i < numRows; ++i) {
            vector<int> row;
            if (!res.empty()) 
                row.assign(res.at(i-1).begin(), res.at(i-1).end());
            row.emplace_back(1);
            for (auto j = i - 1; j > 0; --j)
                row[j] += row[j-1];
            res.emplace_back(row);
        }
        return res;
    }
};

来源:https://github.com/xiaqunfeng/leetcode



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