class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
}
};
64. 最小路径和
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: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
原站题解
python3 解法, 执行用时: 48 ms, 内存消耗: 16.6 MB, 提交时间: 2022-08-20 12:17:32
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]
golang 解法, 执行用时: 8 ms, 内存消耗: 4.3 MB, 提交时间: 2021-08-03 15:07:39
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 }