64. Minimum Path Sum
給定一個 m x n 的網格，裡面填滿了非負數字，找到一條從左上角到右下角的路徑，使沿路徑的所有數字的和最小化。
注意：您只能在任何時候向下或向右移動。
範例 1：
輸入：grid = [[1,3,1],[1,5,1],[4,2,1]] 輸出：7 解釋：因為路徑 1 → 3 → 1 → 1 → 1 最小化了總和。
Python
class Solution:
def minPathSum(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
# Initialize the dp array
dp = [[0] * n for _ in range(m)]
dp[0][0] = grid[0][0]
# Fill the first row
for j in range(1, n):
dp[0][j] = dp[0][j-1] + grid[0][j]
# Fill the first column
for i in range(1, m):
dp[i][0] = dp[i-1][0] + grid[i][0]
# Fill the rest of the dp array
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[m-1][n-1]
18.21MB, 61ms
C++
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
// Initialize the dp array
vector<vector<int>> dp(m, vector<int>(n, 0));
dp[0][0] = grid[0][0];
// Fill the first row
for (int j = 1; j < n; ++j) {
dp[0][j] = dp[0][j-1] + grid[0][j];
}
// Fill the first column
for (int i = 1; i < m; ++i) {
dp[i][0] = dp[i-1][0] + grid[i][0];
}
// Fill the rest of the dp array
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];
}
};
12.65MB, 11ms
Javascript
/**
* @param {number[][]} grid
* @return {number}
*/
var minPathSum = function(grid) {
const m = grid.length;
const n = grid[0].length;
// Initialize the dp array
const dp = Array.from({ length: m }, () => Array(n).fill(0));
dp[0][0] = grid[0][0];
// Fill the first row
for (let j = 1; j < n; j++) {
dp[0][j] = dp[0][j-1] + grid[0][j];
}
// Fill the first column
for (let i = 1; i < m; i++) {
dp[i][0] = dp[i-1][0] + grid[i][0];
}
// Fill the rest of the dp array
for (let i = 1; i < m; i++) {
for (let 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];
};
51.88MB, 58ms
