感覺這是一系列的動態規劃的演算法，正好也將動態規劃的演算法進行一個總結：

演算法一：

帶權重的最小路徑的問題

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.

首先每一個路徑的上一個路徑都是來自於其上方和左方

現將最上面的路徑進行求和，最左邊的路徑進行求和（這裡沒有直接求和的原因是因為用一個一維陣列res[n]記錄路徑，這裡比較巧妙得節約了空間，所以將最左邊的求和演算法放到了迴圈裡面  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;
    }
};
 

和這種動態規劃類似的演算法還有：

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; //<因為只有一種方法
        }
        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];
    }
};

如果只是利用一維空間：

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

這裡涉及到一些障礙物的情況，如果值為1表示這裡有障礙物，不能跨越

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        if(obstacleGrid.empty() || obstacleGrid[0].empty())
        {
            return 0;
        }
        int res[100] = {0};//<還是採用一維陣列
        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)  //<這裡需要稍微注意一下如果當前有障礙表示為0，否則可以延續之前的狀態
                    {
                        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];
    }
};