LC 64. Minimum Path Sum

1. 题目描述

64. Minimum Path Sum

Medium

98225

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.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

2.解题思路

dp数组记录到达（i，j）最短路径。因为题目规定只能向右走或者向下走，dp[ i ][ j ] = min ( dp[ i-1][ j ], dp[ i ][ j-1] ) + (i, j)本身的数值。要注意第一行的元素只能由其左边元素向右走到达，第一列元素只能由其上面元素向下走到达。

3.实现代码

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int row = grid.size(), col = grid[0].size();
        int dp[row][col];
        memset(dp, 0, sizeof(dp));
        dp[0][0] = grid[0][0];
        for (int i=1; i<col; i++) {//根据题意只能向下或向右走，所以第一行只能从（0,0）向右走到达
            dp[0][i] = dp[0][i-1]+grid[0][i];
        }
        for (int i=1; i<row; i++) {
            for (int j=0; j<col; j++) {
                if (j==0)//第一列的元素只能由上一行第一列向下达到
                    dp[i][j] = dp[i-1][j] + grid[i][j];
                else //非第一列元素正常比较哪一条路径更短即可
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
            }
        }
        return dp[row-1][col-1];
    }
};