Leetcode 64. Minimum Path Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/Quincuntial/article/details/84899574

本文详细解析了在网格中寻找从左上角到右下角的最小路径和问题，提供了两种解决方案并附带代码实现。第一种使用额外的二维数组记录路径上的最小和，第二种直接在原网格上进行修改，节省了空间。

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文章作者：Tyan
博客：noahsnail.com | 优快云 | 简书

1. Description

Minimum Path Sum

2. Solution

Version 1

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

Version 2

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