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剑指 Offer II 099. 最小路径之和

题目描述

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：一个机器人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

 

注意：本题与主站 64 题相同： https://leetcode.cn/problems/minimum-path-sum/

解法

方法一

Python3

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]

Java

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];
    }
}

C++

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];
    }
};

Go

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]
}

TypeScript

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];
}

C#

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];
    }
}

Swift

class Solution {
    func minPathSum(_ grid: [[Int]]) -> Int {
        let m = grid.count
        let n = grid[0].count
        var dp = grid

        for i in 1..<m {
            dp[i][0] += dp[i-1][0]
        }

        for j in 1..<n {
            dp[0][j] += dp[0][j-1]
        }

        for i in 1..<m {
            for j in 1..<n {
                dp[i][j] += min(dp[i-1][j], dp[i][j-1])
            }
        }

        return dp[m-1][n-1]
    }
}