64. 最小路径和

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：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] <= 200

解法

方法一：动态规划

我们定义 $f[i][j]$ 表示从左上角走到 $(i, j)$ 位置的最小路径和。初始时 $f[0][0] = grid[0][0]$，答案为 $f[m - 1][n - 1]$。

考虑 $f[i][j]$：

如果 $j = 0$，那么 $f[i][j] = f[i - 1][j] + grid[i][j]$；
如果 $i = 0$，那么 $f[i][j] = f[i][j - 1] + grid[i][j]$；
如果 $i \gt 0$ 且 $j \gt 0$，那么 $f[i][j] = \min(f[i - 1][j], f[i][j - 1]) + grid[i][j]$。

最后返回 $f[m - 1][n - 1]$ 即可。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是网格的行数和列数。

class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        f = [[0] * n for _ in range(m)]
        f[0][0] = grid[0][0]
        for i in range(1, m):
            f[i][0] = f[i - 1][0] + grid[i][0]
        for j in range(1, n):
            f[0][j] = f[0][j - 1] + grid[0][j]
        for i in range(1, m):
            for j in range(1, n):
                f[i][j] = min(f[i - 1][j], f[i][j - 1]) + grid[i][j]
        return f[-1][-1]

class Solution {
    public int minPathSum(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][] f = new int[m][n];
        f[0][0] = grid[0][0];
        for (int i = 1; i < m; ++i) {
            f[i][0] = f[i - 1][0] + grid[i][0];
        }
        for (int j = 1; j < n; ++j) {
            f[0][j] = f[0][j - 1] + grid[0][j];
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
            }
        }
        return f[m - 1][n - 1];
    }
}

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int f[m][n];
        f[0][0] = grid[0][0];
        for (int i = 1; i < m; ++i) {
            f[i][0] = f[i - 1][0] + grid[i][0];
        }
        for (int j = 1; j < n; ++j) {
            f[0][j] = f[0][j - 1] + grid[0][j];
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                f[i][j] = min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
            }
        }
        return f[m - 1][n - 1];
    }
};

func minPathSum(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, n)
	}
	f[0][0] = grid[0][0]
	for i := 1; i < m; i++ {
		f[i][0] = f[i-1][0] + grid[i][0]
	}
	for j := 1; j < n; j++ {
		f[0][j] = f[0][j-1] + grid[0][j]
	}
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			f[i][j] = min(f[i-1][j], f[i][j-1]) + grid[i][j]
		}
	}
	return f[m-1][n-1]
}

function minPathSum(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const f: number[][] = Array(m)
        .fill(0)
        .map(() => Array(n).fill(0));
    f[0][0] = grid[0][0];
    for (let i = 1; i < m; ++i) {
        f[i][0] = f[i - 1][0] + grid[i][0];
    }
    for (let j = 1; j < n; ++j) {
        f[0][j] = f[0][j - 1] + grid[0][j];
    }
    for (let i = 1; i < m; ++i) {
        for (let j = 1; j < n; ++j) {
            f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
        }
    }
    return f[m - 1][n - 1];
}

impl Solution {
    pub fn min_path_sum(mut grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        for i in 1..m {
            grid[i][0] += grid[i - 1][0];
        }
        for i in 1..n {
            grid[0][i] += grid[0][i - 1];
        }
        for i in 1..m {
            for j in 1..n {
                grid[i][j] += grid[i][j - 1].min(grid[i - 1][j]);
            }
        }
        grid[m - 1][n - 1]
    }
}

/**
 * @param {number[][]} grid
 * @return {number}
 */
var minPathSum = function (grid) {
    const m = grid.length;
    const n = grid[0].length;
    const f = Array(m)
        .fill(0)
        .map(() => Array(n).fill(0));
    f[0][0] = grid[0][0];
    for (let i = 1; i < m; ++i) {
        f[i][0] = f[i - 1][0] + grid[i][0];
    }
    for (let j = 1; j < n; ++j) {
        f[0][j] = f[0][j - 1] + grid[0][j];
    }
    for (let i = 1; i < m; ++i) {
        for (let j = 1; j < n; ++j) {
            f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
        }
    }
    return f[m - 1][n - 1];
};

public class Solution {
    public int MinPathSum(int[][] grid) {
        int m = grid.Length, n = grid[0].Length;
        int[,] f = new int[m, n];
        f[0, 0] = grid[0][0];
        for (int i = 1; i < m; ++i) {
            f[i, 0] = f[i - 1, 0] + grid[i][0];
        }
        for (int j = 1; j < n; ++j) {
            f[0, j] = f[0, j - 1] + grid[0][j];
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                f[i, j] = Math.Min(f[i - 1, j], f[i, j - 1]) + grid[i][j];
            }
        }
        return f[m - 1, n - 1];
    }
}

peaking / leetcode

64. 最小路径和

题目描述

解法

方法一：动态规划

简介

发行版

贡献者

近期动态

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64. 最小路径和

题目描述

解法

方法一：动态规划

简介

发行版

贡献者

近期动态

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