64. Minimum Path Sum

问题描述

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.

题目链接：

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思路分析

给一个二维数组，数组中元素大于0，找到一条从左上到右下的最短路径。只能向下或向右移动。

简单的动态规划题，每一个格子距离是min(up, left) + grid[i][j]。计算返回右下角的值返回即可。要注意跳过grid[0][0]，作为递归的开始条件。

代码

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size();
        if (m == 0) return 0;
        int n = grid[0].size();
        for (int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if (i == 0 && j == 0)
                    continue;
                int up = j - 1 >= 0 ? grid[i][j - 1] : INT_MAX;
                int left = i - 1 >= 0 ? grid[i - 1][j] : INT_MAX;
                grid[i][j] = min(up, left) + grid[i][j];
            }
        }
        return grid[m - 1][n - 1];
    }
};

时间复杂度：$O(m+n)$
空间复杂度：$O(1)$

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反思

可以不in-place的做，就需要用一个二维数组dp保存结果值。