剑指 Offer II 099. 最小路径之和
给定一个包含非负整数的 m x n 网格 grid ，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。
说明：一个机器人每次只能向下或者向右移动一步。
示例 1：
输入：grid = [[1,3,1],[1,5,1],[4,2,1]] 输出：7 解释：因为路径 1→3→1→1→1 的总和最小。
示例 2：
输入：grid = [[1,2,3],[4,5,6]] 输出：12
提示：
m == grid.lengthn == grid[i].length1 <= m, n <= 2000 <= grid[i][j] <= 100
注意：本题与主站 64 题相同： https://leetcode.cn/problems/minimum-path-sum/
原站题解
python3 解法, 执行用时: 48 ms, 内存消耗: 16.2 MB, 提交时间: 2022-11-16 11:11:25
class Solution:
def minPathSum(self, grid: [[int]]) -> int:
'''
grid[i][j]原地修改，修改为走到(i, j)位置的最短路径
'''
for i in range(len(grid)):
for j in range(len(grid[0])):
if i == j == 0: continue
elif i == 0: grid[i][j] = grid[i][j - 1] + grid[i][j]
elif j == 0: grid[i][j] = grid[i - 1][j] + grid[i][j]
else: grid[i][j] = min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j]
return grid[-1][-1]
golang 解法, 执行用时: 8 ms, 内存消耗: 4.2 MB, 提交时间: 2022-11-16 11:09:27
func minPathSum(grid [][]int) int {
r, c := len(grid), len(grid[0])
// dp[i][j] 表示向右i, 向下j 的最小和
dp := make([][]int, r)
for i := range dp {
dp[i] = make([]int, c)
}
dp[0][0] = grid[0][0]
for i := 1; i < r; i++ {
dp[i][0] = dp[i - 1][0] + grid[i][0]
}
for j := 1; j < c; j++ {
dp[0][j] = dp[0][j - 1] + grid[0][j]
}
for i := 1; i < r; i++ {
for j := 1; j < c; j++ {
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]
}
}
return dp[r - 1][c - 1]
}
func min(x, y int) int {
if x > y {
return y
}
return x
}
python3 解法, 执行用时: 52 ms, 内存消耗: 16.7 MB, 提交时间: 2022-11-16 11:09:00
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]