【题目】

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.

求从左上角到右下角的最小路径值，典型的动态规划。

动规方程：dp[i][j] = grid[i][j] + min( dp[i-1][j], dp[i][j-1] );

时间复杂度为O(m*n)，因为至少要遍历一遍二维表。

空间复杂度，如果用二维数组来存动规过程，那么就是O(m*n)；但如果用一维数组来存动规过程，空间复杂度就为O(n)。

【Java代码】

public class Solution {
    // DP with two dimension table
    public int minPathSum1(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        
        int[][] ans = new int[m][n];
        ans[0][0] = grid[0][0];
        for (int i = 1; i < m; i++) {
            ans[i][0] = ans[i-1][0] + grid[i][0];
        }
        for (int j = 1; j < n; j++) {
            ans[0][j] = ans[0][j-1] + grid[0][j];
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                ans[i][j] = Math.min(ans[i-1][j], ans[i][j-1]) + grid[i][j];
            }
        }
        
        return ans[m-1][n-1];
    }
    
    // DP with one dimension table
    public int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        
        int[] ans = new int[n];
        ans[0] = grid[0][0];
        for (int j = 1; j < n; j++) {
            ans[j] = ans[j-1] + grid[0][j];
        }
        
        for (int i = 1; i < m; i++) {
            ans[0] += grid[i][0];
            for (int j = 1; j < n; j++) {
                ans[j] = Math.min(ans[j-1], ans[j]) + grid[i][j];
            }
        }
        
        return ans[n-1];
    }
}

【LeetCode】Minimum Path Sum 解题报告