LeetCode - Pascal's Triangle

最新推荐文章于 2025-12-31 20:36:24 发布

原创最新推荐文章于 2025-12-31 20:36:24 发布 · 507 阅读

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文章标签：

#leetcode #array #dp

本文介绍了一种高效的算法来生成帕斯卡三角形。该算法利用动态规划原理，通过简单的数学运算来构建每一层的数值。文章还提供了一个C++实现示例，能够快速生成指定层数的帕斯卡三角形。

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]
]

http://oj.leetcode.com/problems/pascals-triangle/

Solution:

We find that ith row has i numbers. And except the start and end number, t[i][j] = t[i-1][j-1] + t[i-1][j]. So it is a simple one dimension dp problem.

https://github.com/starcroce/leetcode/blob/master/pascal_triangle.cpp

// 4 ms for 15 test cases
class Solution {
public:
    vector<vector<int> > generate(int numRows) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<vector<int> > ans;
        if(numRows == 0) {
            return ans;
        }
        // 1st level
        ans.push_back(vector<int> (1, 1));
        for(int i = 2; i <= numRows; i++) {
            vector<int> level (i, 1);
            // last level
            vector<int> &last = ans[i-2];
            for(int j = 1; j < i-1; j++) {
                level[j] = last[j-1] + last[j];
            }
            ans.push_back(level);
        }
        return ans;
    }
};