63. 不同路径 II @Python@Go

原题

https://leetcode.cn/problems/unique-paths-ii/description/

思路

动态规划

复杂度

时间：O(m * n)
空间：O(m * n)

Python代码

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0] * n for i in range(m)]
        for i in range(m):
            for j in range(n):
                if obstacleGrid[i][j] == 1:
                    dp[i][j] == 0 
                    continue
                elif i == 0 and j == 0:
                    dp[i][j] = 1
                elif i == 0 and j > 0:
                    dp[i][j] = dp[i][j-1]
                elif j == 0 and i > 0:
                    dp[i][j] = dp[i-1][j]
                else:
                    dp[i][j] = dp[i-1][j] + dp[i][j-1]
        return dp[-1][-1]        

Go代码

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
	m, n := len(obstacleGrid), len(obstacleGrid[0])
	// 二维数组储存结果
	dp := make([][]int, m)
	for i := 0; i < m; i++ {
		dp[i] = make([]int, n)
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if obstacleGrid[i][j] == 1 {
				dp[i][j] = 0
				continue
			} else if i == 0 && j == 0 {
				dp[i][j] = 1
			} else if i == 0 && j > 0 {
				dp[i][j] = dp[i][j-1]
			} else if j == 0 && i > 0 {
				dp[i][j] = dp[i-1][j]
			} else {
				dp[i][j] = dp[i-1][j] + dp[i][j-1]
			}
		}
	}
	return dp[m-1][n-1]
}