Question

64. 最小路径和

English Version

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];
  }
}