感觉这是一系列的动态规划的算法，正好也将动态规划的算法进行一个总结：
算法一：
带权重的最小路径的问题
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.
首先每一个路径的上一个路径都是来自于其上方和左方
现将最上面的路径进行求和，最左边的路径进行求和（这里没有直接求和的原因是因为用一个一维数组res[n]记录路径，这里比较巧妙得节约了空间，所以将最左边的求和算法放到了循环里面 res[i][j] = min(res[i-1][j],res[i][j-1])+val[i][j]
）
class Solution {
public:
int minPathSum(vector<vector<int> > &grid) {
if(grid.empty() || grid[0].empty())
{
return 0;
}
int m = grid.size();
int n = grid[0].size();
int *res = new int[n];
res[0] = grid[0][0];
for(int i = 1;i < n; i++)
{
res[i] = res[i-1]+grid[0][i];
}
for(int i = 1; i < m; i++)
{
for(int j = 0; j < n; j++)
{
if (0 == j)
{
res[j] += grid[i][0];
}
else
{
res[j] = min(res[j-1],res[j])+grid[i][j];
}
}
}
int result = res[n-1];
delete []res;
return result;
}
};
和这种动态规划类似的算法还有：
Unique Paths
https://leetcode.com/problems/unique-paths/
A robot is located at the top-left corner of a m x n grid (marked ‘Start‘ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish‘ in the diagram below).
How many possible unique paths are there?
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
class Solution {
public:
int uniquePaths(int m, int n) {
if(m == 0 || n == 0)
{
return 0;
}
int res[100][100];
for(int i = 0; i < n; i++)
{
res[0][i] = 1; //<因为只有一种方法
}
for(int i = 0; i < m; i++)
{
res[i][0] = 1;
}
for(int i = 1; i < m; i++)
{
for(int j = 1; j < n; j++)
{
res[i][j] = res[i-1][j]+res[i][j-1];
}
}
return res[m-1][n-1];
}
};
如果只是利用一维空间：
class Solution {
public:
int uniquePaths(int m, int n) {
if(m <= 0 || n <= 0
{
return 0;
}
int res[100] = {0};
res[0] = 1;
for(int i = 0; i < m; i++)
{
for(int j = 1; j < n; j++)
{
res[j] += res[j-1];
}
}
return res[n-1];
}
};
Unique Paths II
这里涉及到一些障碍物的情况，如果值为1表示这里有障碍物，不能跨越
[ [0,0,0], [0,1,0], [0,0,0] ]
The total number of unique paths is 2.
Note: m and n will
be at most 100.
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
if(obstacleGrid.empty() || obstacleGrid[0].empty())
{
return 0;
}
int res[100] = {0};//<还是采用一维数组
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
res[0] = (obstacleGrid[0][0] != 1);
for(int i = 0; i < m; i++)
{
for(int j = 0; j < n; j++)
{
<span style="color:#ff0000;">if(j == 0)
{
if(obstacleGrid[i][j] == 1) //<这里需要稍微注意一下如果当前有障碍表示为0，否则可以延续之前的状态
{
res[j] = 0;
}
}</span>
else
{
<span style="color:#ff0000;">if(obstacleGrid[i][j] == 1)
{
res[j] = 0;
}
else
{
res[j] += res[j-1];
}</span>
}
}
}
return res[n-1];
}
};