【leetcode】Minimum Path Sum

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.

这道题目明显用动态规划，最优子结构为

a[i][j]=min(a[i][j-1],a[i-1][j])+grid[i][j];

注意
（1）本题我用malloc来开辟二维数组，开辟的主要过程为

int **a = (int **)malloc(sizeof(int *)*(m+1));  for(int i=0;i<(m+1);i++)     {          a[i] = (int *)malloc(sizeof(int)*(n+1));     }

但是在二维数组初始化的过程中，不能简简单单用memset:

for (int i = 0; i < m+1; ++i)    memset(a[i], -1, sizeof(int)*(n+1));

（2）边界条件的书写

class Solution {public:    int minPathSum(vector<vector<int> > &grid) {        int m=grid.size();        int n=grid[0].size();        if(m==0 || n==0)return 0;        //int a[1000][1000];        //memset(a, -1, sizeof(a));        int **a = (int **)malloc(sizeof(int *)*(m+1));          for(int i=0;i<(m+1);i++)          {              a[i] = (int *)malloc(sizeof(int)*(n+1));         }        for (int i = 0; i < m+1; ++i)            memset(a[i], -1, sizeof(int)*(n+1));        //边界        a[0][0]=grid[0][0];        for(int i=1;i<m;i++)            a[i][0]=a[i-1][0]+grid[i][0];        for(int j=1;j<n;j++)            a[0][j]=a[0][j-1]+grid[0][j];        for(int i=1;i<m;i++)        {            for(int j=1;j<n;j++)            {                a[i][j]=min(a[i][j-1],a[i-1][j])+grid[i][j];            }        }        return a[m-1][n-1];    }};