2022.03.18-第49题-64. 最小路径和

[icodeish]

https://leetcode-cn.com/problems/minimum-path-sum/ 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

[icodeish]

function minPathSum(grid) {
    for (let i = 1; i < grid.length; i++) grid[i][0] += grid[i - 1][0]
    for (let j = 1; j < grid[0].length; j++) grid[0][j] += grid[0][j - 1]
    for (let i = 1; i < grid.length; i++)
        for (let j = 1; j < grid[0].length; j++)
            grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1])
    return grid[grid.length - 1][grid[0].length - 1]
}

[BambooSword]

function minPathSum(grid: number[][]): number {
const dp = new Array(grid.length)
    for (let i = 0; i < grid.length; ++i) {
        dp[i] = []
        for (let j = 0; j < grid[i].length; ++j) {
            if (i === 0 && j === 0) dp[i][j] = grid[i][j]
            else {
                if (i === 0) dp[i][j] = dp[i][j - 1] + grid[i][j]
                else if (j === 0) dp[i][j] = dp[i - 1][j] + grid[i][j]
                else dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]
            }
        }
    }

    const m = dp.length - 1
    const n = dp[0].length - 1

    return dp[m][n]
};

[yanggengzhen123]

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