参考链接
题目描述
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.
题目分析
就是给一个m*n的矩阵,从左上角到右下角,找出一条通过数字之和最小的路。
用动态规划:path[i][j] = MIN(path[i][j-1],path[i-1][j])+num[i][j];
代码示例
/*
编译环境CFree 5.0
博客地址:http://blog.youkuaiyun.com/Snowwolf_Yang
*/
#include
#include
using namespace std;
void printvec(vector
A)
{
for(int i = 0;i
> A, char *name)
{
printf("%s\n",name);
for(int i = 0;i
> &grid) {
int size = grid.size();
if(size == 0) return 0;
vector
> minPath(size,vector
(grid[0].size())); minPath[0][0] = grid[0][0]; for(int i = 1;i
> grid; vector
tmp; for(int i=0;i

本文详细解析了如何求解从矩阵左上角到右下角的最小路径和问题,采用动态规划方法,提供了完整的代码实现,并附带推荐的学习资源。
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