[LeetCode] 64. Minimum Path Sum（最小路径和）

• Difficulty: Medium

• Related Topics: Array, Dynamic Programming

• Link: https://leetcode.com/problems/minimum-path-sum/

Description

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.
给定一个 m x n 的矩阵，矩阵元素都是非负书，寻找一条从左上角到右下角的路径，使得路径上的数的和最小。

Note

You can only move either down or right at any point in time.

Examples

Example 1

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

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

Solution

动态规划其中一道例题，状态方程如下：

[ exttt{dp(i, j)} = egin{cases} exttt{dp(i - 1, j)} + exttt{grid[i][j]}, & j = 0 \ exttt{dp(i, j - 1)} + exttt{grid[i][j]}, & i = 0 \ exttt{dp(i - 1, j - 1)} + min( exttt{grid[i - 1][j]}, exttt{grid[i][j - 1]}), & else end{cases} ]

其中 dp(i, j) 表示到 (i, j) 位置时的最短路径

import kotlin.math.min

class Solution {
    fun minPathSum(grid: Array<IntArray>): Int {
        val dp = Array(grid.size) { IntArray(grid[0].size) }
        dp[0][0] = grid[0][0]
        for (i in 1..grid.lastIndex) {
            dp[i][0] = dp[i - 1][0] + grid[i][0]
        }
        for (i in 1..grid[0].lastIndex) {
            dp[0][i] = dp[0][i - 1] + grid[0][i]
        }

        for (i in 1..grid.lastIndex) {
            for (j in 1..grid[i].lastIndex) {
                dp[i][j] = grid[i][j] + min(dp[i - 1][j], dp[i][j - 1])
            }
        }

        return dp.last().last()
    }
}