每日一题 day 50(DP topic)

最新推荐文章于 2025-12-08 22:55:11 发布

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#动态规划 #leetcode #算法

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本文探讨了如何使用动态规划(DP)解决最小路径和问题，包括常规DP方法和空间优化版本。通过实例分析，展示了两种方法在求解64x64网格中从左上角到右下角的最短路径总和时的应用。第一种方法使用二维数组，而第二种方法仅用一维数组，以减少空间复杂度。

文章目录

problem
approach 1`DP`
approach 2 `DP less space`

problem

64. Minimum Path Sum
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 1:
在这里插入图片描述

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

Example 2:

Input: grid = [[1,2,3],[4,5,6]]
Output: 12

approach 1`DP`

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<int>> dp(m+1, vector<int>(n+1, 999));
        dp[0][1] = 0;
        for(int i=1; i<=m; i++){
            for(int j=1; j<=n; j++){
                dp[i][j] = grid[i-1][j-1] + min(dp[i-1][j], dp[i][j-1]);
            }
        }
        return dp[m][n];
    }
};

approach 2 `DP less space`

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