64. Minimum Path Sum

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.

解题思路为Dynamic Programming。

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

可以把space complexity缩减为O(n)

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