leetcode-64-Minimum-Path-Sum

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描述

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 1:

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Input: grid = [[1,3,1],[1,5,1],[4,2,1]]
Output: 7
Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:

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Input: grid = [[1,2,3],[4,5,6]]
Output: 12

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 200
  • 0 <= grid[i][j] <= 100

分析

动态规划。推导公式是：grid[i][j] = min(grid[i-1][j], grid[i][j-1]) + grid[i][j], 注意处理边界情况即可。

解决方案1(Python3)

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class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if i == j == 0:
                    continue
                elif i == 0:
                    grid[i][j] = grid[i][j-1] + grid[i][j]
                elif j == 0:
                    grid[i][j] = grid[i-1][j] + grid[i][j]
                else:
                    grid[i][j] = min(grid[i-1][j], grid[i][j-1]) + grid[i][j]
        return grid[-1][-1]

解决方案2（Golang）

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func minPathSum(grid [][]int) int {
    if len(grid) == 0 || len(grid[0]) == 0 {
        return 0
    }
    rows, columns := len(grid), len(grid[0])
    dp := make([][]int, rows)
    for i := 0; i < len(dp); i++ {
        dp[i] = make([]int, columns)
    }
    dp[0][0] = grid[0][0]
    
    for i := 1; i < rows; i++ {
        dp[i][0] = dp[i-1][0] + grid[i][0]
    }
    for j := 1; j < columns; j++ {
        dp[0][j] = dp[0][j-1] + grid[0][j]
    }
    
    for i := 1; i < rows; i++ {
        for j := 1; j < columns; j++ {
            dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
        }
    }
    return dp[rows-1][columns-1]
}

func min(x, y int) int {
    if x < y {
        return x
    }
    return y
}

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