64. Minimum Path Sum

problem description

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.

algorithm thought

和上一题没什么不同，动态规划换汤不换药。上一题是路线的数量，这里是找到最小值。

code

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

algorithm analysis

时间复杂度和上两题一样,O(n²)