【Leetcode长征系列】Minimum Path Sum

原题：

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right whichminimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

思路：看到最短路径就会脑补到图论的算法。但是本题并不需要那么复杂的动态规划。因为它本身是一个矩阵的结构，在每一个节点的值都不为负的情况下，我们只需要找到下边和右边方向中较小的那个值再相加即可。这样局部最优的方式来找到整体最优。算基本的动态规划题。

class Solution {
public:
int minPathSum(vector<vector<int> > &grid) {
int dp[1000][1000];
int col = grid.size();
int row = grid[0].size();
if (grid.size() == 0 || grid[0].size() == 0)     return 0;

for(int i = 0; i < row; i++){
for(int j = 0; j < col; j++){
if(i==0 && j==0) dp[i][j] = grid[i][j];
else if(i==0 && j > 0) dp[i][j] = grid[i][j-1]+grid[i][j];
else if(i > 0 && j==0) dp[i][j] = grid[i-1][j]+grid[i][j];
else dp[i][j] = min(dp[i-1][j],dp[i][j-1])+grid[i][j];
}
}
return dp[row-1][col-1];
}
};

我真得已经debug不出来了QAQ

网上找了一份代码：

int dp[1000][1000];
class Solution {
public:
int minPathSum(vector<vector<int> > &grid) {
int rows=grid.size();
if(rows==0)return 0;
int cols=grid[0].size();
if(cols==0)return 0;
memset(dp,0,sizeof(dp));
for(int i=0;i<rows;++i)
{
for(int j=0;j<cols;++j)
{
if(i==0&&j==0)dp[0][0]=grid[0][0];
else if(j==0)dp[i][0]=dp[i-1][0]+grid[i][0];
else if(i==0)dp[0][j]=dp[0][j-1]+grid[0][j];
else dp[i][j]=min(dp[i-1][j],dp[i][j-1])+grid[i][j];
}
}
return dp[rows-1][cols-1];
}
};

what's the difference?!这个就AC了T________________T



===========================================================================================================================

嗯在被喷两次以后

class Solution {
public:
int minPathSum(vector<vector<int> > &grid) {
int dp[1000][1000];
int row = grid.size();
int col = grid[0].size();
if (grid.size() == 0 || grid[0].size() == 0) return 0;

for(int i = 0; i < row; i++){
for(int j = 0; j < col; j++){
if(i==0&&j==0) dp[0][0]=grid[0][0];
else if(j==0) dp[i][0]=dp[i-1][0]+grid[i][0];
else if(i==0) dp[0][j]=dp[0][j-1]+grid[0][j];
else dp[i][j]=min(dp[i-1][j],dp[i][j-1])+grid[i][j];
}
}
return dp[row-1][col-1];
}
};


为啥算法一样这次的快了那么多0 0