Leetcode64. 最小路径和

Problem: 64. 最小路径和

文章目录

• 思路
• 解题方法
• 复杂度
• Code

思路

动态规划,偷房子问题变形

解题方法

dpij = min(dpi-1j,dpij-1)+gridij;

复杂度

时间复杂度:

O ( m ∗ n ) O(m*n) O(m∗n)

空间复杂度:

O ( m ∗ n ) O(m*n) O(m∗n)

Code

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

        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                dp[i][j] = min(dp[i-1][j],dp[i][j-1])+grid[i][j];
            }
        }
        return dp[m - 1][n - 1];
    }
};