Question

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

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]

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];
  }
}

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];
  }
};

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]
}

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];
}

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];
  }
}