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：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.length
- n == grid[i].length
- 1 <= m, n <= 200
- 0 <= grid[i][j] <= 100
解题：
func minPathSum(grid [][]int) int {
m, n := len(grid), len(grid[0])
r := make([][]int, m)
for i := 0; i < m; i++ {
r[i] = make([]int, n)
if i == 0 {
r[i][0] = grid[0][0]
for j := 1; j < n; j++ {
r[i][j] = r[i][j-1] + grid[i][j]
}
} else {
r[i][0] = r[i-1][0] + grid[i][0]
}
}
for i := 1; i < m; i++ {
for j := 1; j < n; j++ {
r[i][j] = min(r[i-1][j], r[i][j-1]) + grid[i][j]
}
}
return r[m-1][n-1]
}
func min(a, b int) int {
if a < b {
return a
}
return b
}官方解答：
1.动态规划
func minPathSum(grid [][]int) int {
if len(grid) == 0 || len(grid[0]) == 0 {
return 0
}
rows, columns := len(grid), len(grid[0])
dp := make([][]int, rows)
for i := 0; i < len(dp); i++ {
dp[i] = make([]int, columns)
}
dp[0][0] = grid[0][0]
for i := 1; i < rows; i++ {
dp[i][0] = dp[i-1][0] + grid[i][0]
}
for j := 1; j < columns; j++ {
dp[0][j] = dp[0][j-1] + grid[0][j]
}
for i := 1; i < rows; i++ {
for j := 1; j < columns; j++ {
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
}
}
return dp[rows-1][columns-1]
}
func min(x, y int) int {
if x < y {
return x
}
return y
}