13.8 Minimum Path Sum

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

My thought: Simple 2d-DP. I passed OJ on my first attempt.

Approach I : Time: O(mn), Space: O(1)

public class Solution {    //my 1st attempt: g[i][j] = min(g[i-1][j], g[i][j-1]) + g[i][j]    public int minPathSum(int[][] grid) {        if(grid == null) return 0;        //init the 1st row        for(int j = 1; j < grid[0].length; j++){            grid[0][j] += grid[0][j-1];        }        //init the 1st col        for(int i = 1; i < grid.length; i++){            grid[i][0] += grid[i-1][0];        }        for(int i = 1; i < grid.length; i++){            for(int 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];    }}

Approach II: DP with 滚动数组

Time: O(mn), Space: O(n)

public class Solution {    //my 2nd attempt    public int minPathSum(int[][] grid) {        if(grid == null || grid.length == 0 || grid[0].length == 0) return 0;        int[] res = new int[grid[0].length];                //init the 1st row: i = 0        res[0] = grid[0][0];        for(int j = 1; j < grid[0].length; j++){            res[j] = res[j-1] + grid[0][j];        }        for(int i = 1; i < grid.length; i++){            res[0] += grid[i][0];            for(int j = 1; j < grid[0].length; j++){                res[j] = Math.min(res[j], res[j-1]) + grid[i][j];            }        }        return res[grid[0].length-1];    }}

相关题目：10.2 Unique Paths,10.3 Unique Paths II

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