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118. Pascal’s Triangle

Description

Given an integer numRows, return the first numRows of Pascal's triangle.

In Pascal's triangle, each number is the sum of the two numbers directly above it as shown:

 

Example 1:

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

Example 2:

Input: numRows = 1
Output: [[1]]

 

Constraints:

• 1 <= numRows <= 30

Solutions

Solution 1: Simulation

First, we create an answer array $f$, and then set the first row of $f$ to $[1]$. Next, starting from the second row, the first and last elements of each row are $1$, and the other elements are calculated by $f[i][j] = f[i - 1][j - 1] + f[i - 1][j]$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the number of rows.

• Java
• C++
• Python
• Go
• TypeScript
• Javascript
• RenderScript

• class Solution {
      public List<List<Integer>> generate(int numRows) {
          List<List<Integer>> f = new ArrayList<>();
          f.add(List.of(1));
          for (int i = 0; i < numRows - 1; ++i) {
              List<Integer> g = new ArrayList<>();
              g.add(1);
              for (int j = 0; j < f.get(i).size() - 1; ++j) {
                  g.add(f.get(i).get(j) + f.get(i).get(j + 1));
              }
              g.add(1);
              f.add(g);
          }
          return f;
      }
  }
  

• class Solution {
  public:
      vector<vector<int>> generate(int numRows) {
          vector<vector<int>> f;
          f.push_back(vector<int>(1, 1));
          for (int i = 0; i < numRows - 1; ++i) {
              vector<int> g;
              g.push_back(1);
              for (int j = 0; j < f[i].size() - 1; ++j) {
                  g.push_back(f[i][j] + f[i][j + 1]);
              }
              g.push_back(1);
              f.push_back(g);
          }
          return f;
      }
  };
  

• class Solution:
      def generate(self, numRows: int) -> List[List[int]]:
          f = [[1]]
          for i in range(numRows - 1):
              g = [1] + [a + b for a, b in pairwise(f[-1])] + [1]
              f.append(g)
          return f
  
  

• func generate(numRows int) [][]int {
  	f := [][]int{[]int{1} }
  	for i := 0; i < numRows-1; i++ {
  		g := []int{1}
  		for j := 0; j < len(f[i])-1; j++ {
  			g = append(g, f[i][j]+f[i][j+1])
  		}
  		g = append(g, 1)
  		f = append(f, g)
  	}
  	return f
  }
  

• function generate(numRows: number): number[][] {
      const f: number[][] = [[1]];
      for (let i = 0; i < numRows - 1; ++i) {
          const g: number[] = [1];
          for (let j = 0; j < f[i].length - 1; ++j) {
              g.push(f[i][j] + f[i][j + 1]);
          }
          g.push(1);
          f.push(g);
      }
      return f;
  }
  
  

• /**
   * @param {number} numRows
   * @return {number[][]}
   */
  var generate = function (numRows) {
      const f = [[1]];
      for (let i = 0; i < numRows - 1; ++i) {
          const g = [1];
          for (let j = 0; j < f[i].length - 1; ++j) {
              g.push(f[i][j] + f[i][j + 1]);
          }
          g.push(1);
          f.push(g);
      }
      return f;
  };
  
  

• impl Solution {
      #[allow(dead_code)]
      pub fn generate(num_rows: i32) -> Vec<Vec<i32>> {
          let mut ret_vec: Vec<Vec<i32>> = Vec::new();
          let mut cur_vec: Vec<i32> = Vec::new();
  
          for i in 0..num_rows as usize {
              let mut new_vec: Vec<i32> = vec![1; i + 1];
              for j in 1..i {
                  new_vec[j] = cur_vec[j - 1] + cur_vec[j];
              }
              cur_vec = new_vec.clone();
              ret_vec.push(new_vec);
          }
  
          ret_vec
      }
  }
  
  

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