[Leetcode]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:

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

题目解释：就是从数组的左上角走到数组右下角的最短路径
解法一：正向

class Solution {
public:
    int minPathSum(vector>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int dip[m][n];   // c++定义数组的方式,dip数组中存放grid中相应位置的最短路径
        dip[0][0] = grid[0][0];
        for(int i = 1; i < m; i++) dip[i][0] = grid[i][0]+dip[i-1][0];//把上边缘放入
        for(int j = 1; j < n; j++) dip[0][j] = grid[0][j]+dip[0][j-1];//把左边缘放入
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                dip[i][j] = min(dip[i-1][j],dip[i][j-1])+grid[i][j];
            }
        }
        return dip[m-1][n-1];
    }
}

解法二：递归，逆向，超时

class Solution {
public:
    int minPathSum(vector>& grid) {
         return funn(grid,grid.size()-1,grid[0].size()-1);
    }
    // 递归 超时
    int funn(vector>& grid,int row,int col){         
        if(row == 0 &&  col == 0) return grid[0][0];// 防止到grid[0][0]处返回的是INT_MAX；
        if(row >= 0 && row < grid.size()&& col >= 0 && col < grid[0].size()){
            int leftmin = funn(grid,row,col-1); // 左二维数组最小和
            int rightmin = funn(grid,row-1,col); // 右二维数组最小和
            return min(leftmin,rightmin)+grid[row][col];
        }
        return INT_MAX;
    }
};