64 Minimum Path Sum 最小路径和
Description:
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note:
You can only move either down or right at any point in time.
Example:
Input:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.
题目描述:
给定一个包含非负整数的 m x n 网格，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 :
输入:
[
[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。
思路:
动态规划
- 边界 dp[i][0] = dp[i][0] + dp[i - 1][0], dp[0][i] = dp[0][i] + dp[0][i - 1]
- 内部 dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + dp[i][j]
时间复杂度O(mn), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int minPathSum(vector<vector<int>>& grid)
{
if (grid.empty()) return 0;
for (int i = 1; i < grid.size(); i++) grid[i][0] += grid[i - 1][0];
for (int i = 1; i < grid[0].size(); i++) grid[0][i] += grid[0][i - 1];
for (int i = 1; i < grid.size(); i++) for (int j = 1; j < grid[0].size(); j++) grid[i][j] += min(grid[i - 1][j], grid[i][j - 1]);
return grid.back().back();
}
};
Java:
class Solution {
public int minPathSum(int[][] grid) {
if (grid.length == 0 || grid[0].length == 0) return 0;
for (int i = 1; i < grid.length; i++) grid[i][0] += grid[i - 1][0];
for (int i = 1; i < grid[0].length; i++) grid[0][i] += grid[0][i - 1];
for (int i = 1; i < grid.length; i++) for (int j = 1; j < grid[0].length; j++) grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1]);
return grid[grid.length - 1][grid[0].length - 1];
}
}
Python:
class Solution:
def minPathSum(self, grid: List[List[int]]) -> int:
if not grid or not grid[0]:
return 0
for i in range(1, len(grid)):
grid[i][0] += grid[i - 1][0]
for j in range(1, len(grid[0])):
grid[0][j] += grid[0][j - 1]
for i in range(1, len(grid)):
for j in range(1, len(grid[0])):
grid[i][j] += min(grid[i - 1][j], grid[i][j - 1])
return grid[-1][-1]
