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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
动态规划。假设 dp[i][j]
表示到达网格 (i,j)
的最小数字和,先初始化 dp 第一列和第一行的所有值,然后利用递推公式:dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]
求得 dp。
最后返回 dp[m - 1][n - 1]
即可。
class Solution:
def minPathSum(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
dp = [[grid[0][0]] * n for _ in range(m)]
for i in range(1, m):
dp[i][0] = dp[i - 1][0] + grid[i][0]
for j in range(1, n):
dp[0][j] = dp[0][j - 1] + grid[0][j]
for i in range(1, m):
for j in range(1, n):
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]
return dp[-1][-1]
class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[][] dp = new int[m][n];
dp[0][0] = grid[0][0];
for (int i = 1; i < m; ++i) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
for (int j = 1; j < n; ++j) {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[m - 1][n - 1];
}
}
function minPathSum(grid: number[][]): number {
let m = grid.length,
n = grid[0].length;
let dp = Array.from({ length: m }, v => new Array(n).fill(0));
dp[0][0] = grid[0][0];
for (let i = 1; i < m; ++i) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
for (let j = 1; j < n; ++j) {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
// dp
for (let i = 1; i < m; ++i) {
for (let j = 1; j < n; ++j) {
let cur = grid[i][j];
dp[i][j] = cur + Math.min(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[m - 1][n - 1];
}
class Solution {
public:
int minPathSum(vector<vector<int>> &grid) {
int m = grid.size(), n = grid[0].size();
vector<vector<int>> dp(m, vector<int>(n, grid[0][0]));
for (int i = 1; i < m; ++i)
{
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
for (int j = 1; j < n; ++j)
{
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
for (int i = 1; i < m; ++i)
{
for (int j = 1; j < n; ++j)
{
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[m - 1][n - 1];
}
};
func minPathSum(grid [][]int) int {
m, n := len(grid), len(grid[0])
dp := make([][]int, m)
for i := 0; i < m; i++ {
dp[i] = make([]int, n)
}
dp[0][0] = grid[0][0]
for i := 1; i < m; i++ {
dp[i][0] = dp[i-1][0] + grid[i][0]
}
for j := 1; j < n; j++ {
dp[0][j] = dp[0][j-1] + grid[0][j]
}
for i := 1; i < m; i++ {
for j := 1; j < n; j++ {
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
}
}
return dp[m-1][n-1]
}
func min(a, b int) int {
if a < b {
return a
}
return b
}
public class Solution {
public int MinPathSum(int[][] grid) {
int m = grid.Length, n = grid[0].Length;
int[,] dp = new int[m, n];
dp[0, 0] = grid[0][0];
for (int i = 1; i < m; ++i)
{
dp[i, 0] = dp[i - 1, 0] + grid[i][0];
}
for (int j = 1; j < n; ++j)
{
dp[0, j] = dp[0, j - 1] + grid[0][j];
}
for (int i = 1; i < m; ++i)
{
for (int j = 1; j < n; ++j)
{
dp[i, j] = Math.Min(dp[i - 1, j], dp[i, j - 1]) + grid[i][j];
}
}
return dp[m- 1, n - 1];
}
}
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