原题
https://leetcode.cn/problems/unique-paths/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 uniquePaths(self, m: int, n: int) -> int:
dp = [[0] * n for i in range(m)]
dp[0] = [1] * n
if m == 1:
return dp[0][-1]
for i in range(1, m):
for j in range(n):
if j == 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 uniquePaths(m int, n int) int {
// 创建二维数组, 储存结果
dp := make([][]int, m)
for i := 0; i < m; i++ {
dp[i] = make([]int, n)
}
for j := 0; j < n; j++ {
dp[0][j] = 1
}
if m == 1 {
return dp[0][n-1]
}
for i := 1; i < m; i++ {
for j := 0; j < n; j++ {
if j == 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]
}
108

被折叠的 条评论
为什么被折叠?



