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.

You can only go right or down in the path..

Example

Example 1:
	Input:  [[1,3,1],[1,5,1],[4,2,1]]
	Output: 7
	
	Explanation:
	Path is: 1 -> 3 -> 1 -> 1 -> 1


Example 2:
	Input:  [[1,3,2]]
	Output: 6
	
	Explanation:  
	Path is: 1 -> 3 -> 2

思路：Dp[i][j] 存储从（0， 0） 到（i, j）的最短路径。
Dp[i][j] = min(Dp[i-1][j]), Dp[i][j-1]) + grid[i][j];

public class Solution {
    /**
     * @param grid: a list of lists of integers
     * @return: An integer, minimizes the sum of all numbers along its path
     */
    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[][] sum = new int[M][N];

        sum[0][0] = grid[0][0];

        for (int i = 1; i < M; i++) {
            sum[i][0] = sum[i - 1][0] + grid[i][0];
        }

        for (int i = 1; i < N; i++) {
            sum[0][i] = sum[0][i - 1] + grid[0][i];
        }

        for (int i = 1; i < M; i++) {
            for (int j = 1; j < N; j++) {
                sum[i][j] = Math.min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];
            }
        }

        return sum[M - 1][N - 1];
    }
    
}

　　可以使用滚动数组进行优化。

public class Solution {
    /**
     * @param grid: a list of lists of integers
     * @return: An integer, minimizes the sum of all numbers along its path
     */
     public int minPathSum(int[][] A) {
        if (A == null || A.length == 0 || A[0].length == 0) {
            return 0;
        }
        
        int m = A.length, n = A[0].length;
        int[][] f = new int[2][n];
        int i, j;
        int old, now = 0; // f[i] is stored in rolling array f[0]
        for (i = 0; i < m; ++i) {
            old = now;
            now = 1 - now; // 0 --> 1, 1 --> 0
            
            for (j = 0; j < n; ++j) {
                if (i == 0 && j == 0) {
                    f[now][j] = A[0][0];
                    continue;
                }
                
                f[now][j] = Integer.MAX_VALUE;
                if (i > 0) {
                    f[now][j] = Math.min(f[now][j], f[old][j]);
                }
                
                if (j > 0) {
                    f[now][j] = Math.min(f[now][j], f[now][j - 1]);
                }
                
                f[now][j] += A[i][j];
            }
        }
        
        return f[now][n - 1];
    }
    
}