letcode 矩阵-day

118. 杨辉三角

class Solution:
    def generate(self, numRows: int) -> List[List[int]]:
        ret = list()
        for i in range(numRows):
            row = list()
            for j in range(0, i + 1):
                if j == 0 or j == i:
                    row.append(1)
                else:
                    row.append(ret[i - 1][j] + ret[i - 1][j - 1])
            ret.append(row)
        return ret

64. 最小路径和

https://leetcode.cn/problems/minimum-path-sum/solution/zui-xiao-lu-jing-he-by-leetcode-solution/

class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        if not grid or not grid[0]:
            return 0
        
        rows, columns = len(grid), len(grid[0])
        dp = [[0] * columns for _ in range(rows)]
        dp[0][0] = grid[0][0]
        for i in range(1, rows):
            dp[i][0] = dp[i - 1][0] + grid[i][0]
        for j in range(1, columns):
            dp[0][j] = dp[0][j - 1] + grid[0][j]
        for i in range(1, rows):
            for j in range(1, columns):
                dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]
        
        return dp[rows - 1][columns - 1]

面试题 01.07. 旋转矩阵

给你一幅由 N × N 矩阵表示的图像，其中每个像素的大小为 4 字节。请你设计一种算法，将图像旋转 90 度。

class Solution {
    public void rotate(int[][] matrix) {
        int row = matrix.length;
        int col = matrix[0].length;
        int[][] nums = new int[row][col];
        for(int i = 0 ; i < row ; i++){
            for(int j = 0 ; j < col ; j++){
                nums[j][col - i - 1] = matrix[i][j];
            }
        }
        for(int i = 0 ; i < row ; i++){
            for(int j = 0 ; j < col ; j++){
                matrix[i][j] = nums[i][j];
            }
        }
    }
}