LeetCode 64. Minimum Path Sum

题目如下：

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.

思路：

因为只能往右或往下走，所以可以初始化当res[m][0], res[0]
这些边界情况

然后，每一步结果为当前值＋min(上边一步，左边一步)，即res[i][j] = grid[i][j] + min(res[i-1][j], res[i][j-1])

本体代码：

/**
* Created by yuanxu on 17/4/13.
*/
public class DP64 {

public static int minPathSum(int[][] grid) {
int m = grid.length;
int n = grid[0].length;
int res[][] = new int[m]
;

// boundary conditions
res[0][0] = grid[0][0];
for (int i=1; i<m; i++) {
res[i][0] = grid[i][0] + res[i-1][0];
}
for (int j=1; j<n; j++) {
res[0][j] = grid[0][j] + res[0][j-1];
}

// dp
for (int i=1; i<m; i++) {
for (int j=1; j<n; j++) {
res[i][j] = grid[i][j] + (res[i-1][j] < res[i][j-1] ? res[i-1][j] : res[i][j-1]);
}
}

return res[m-1][n-1];
}

public  static void main(String[] args) {
int grid[][] = {{1,2},{1,1}};
System.out.println(minPathSum(grid));
}
}

网上的解答有空间复杂度为0的：

public class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length;// row
int n = grid[0].length; // column
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 && j != 0) {
grid[i][j] = grid[i][j] + grid[i][j - 1];
} else if (i != 0 && j == 0) {
grid[i][j] = grid[i][j] + grid[i - 1][j];
} else if (i == 0 && j == 0) {
grid[i][j] = grid[i][j];
} else {
grid[i][j] = Math.min(grid[i][j - 1], grid[i - 1][j])
+ grid[i][j];
}
}
}

return grid[m - 1][n - 1];
}
}

ref: https://discuss.leetcode.com/topic/5459/my-java-solution-using-dp-and-no-extra-space