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.

二维动态规划

说明

state: dpx从起点走到x,y的最短路径
function: dpx = min(dpx-1, dpx) + costx
intialize: dp0 = cost0
// fi = sum(0,0 -> i,0)
// f0 = sum(0,0 -> 0,i)
answer: fn-1

复杂度

时间O(mn) 空间O(mn)

代码

public int minPathSum(int[][] grid) {    int m = grid.length;    int n = grid[0].length;    int[][] dp = new int[m][n];    dp[0][0] = grid[0][0];    for (int i = 1; i < n; i++) {        dp[0][i] = grid[0][i] + dp[0][i - 1];    }    for (int j = 1; j < m; j++) {        dp[j][0] = grid[j][0] + dp[j - 1][0];    }    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];}

一维动态规划

说明

初始化第一行的数据, 然后每次遍历一行覆盖一次数据, 思路与二维相同

复杂度

时间O(m*n) 空间O(n)

代码

public int minPathSum(int[][] grid) {    int m = grid.length;    int n = grid[0].length;    int[] dp = new int[n];    dp[0] = grid[0][0];    for (int i = 1; i < n; i++) {        dp[i] = dp[i - 1] + grid[0][i];    }    for (int i = 1; i < m; i++) {        for (int j = 0; j < n; j++) {if (j != 0) {    dp[j] = Math.min(dp[j], dp[j - 1]);}dp[j] += grid[i][j];        }    }    return dp[n - 1];}