[LeetCode] Unique Paths、Unique Paths II、Minimum Path Sum

最新推荐文章于 2023-08-03 01:42:09 发布

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最新推荐文章于 2023-08-03 01:42:09 发布

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本文探讨了在给定网格中寻找从起点到终点的不同路径数量及最小路径和的问题，并考虑了障碍物的存在。通过动态规划方法实现了路径计数和路径求和。

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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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        vector<vector<int> > grid(m,vector<int>(n,0));
	    for(int i=0;i<m;i++)
		    grid[i][0]=1;
	    for(int i=0;i<n;i++)
		    grid[0][i]=1;
	    for(int i=1;i<m;i++)
		    for(int j=1;j<n;j++)
			    grid[i][j]=grid[i-1][j]+grid[i][j-1];
	    return grid[m-1][n-1];
    }
};

Unique Paths II:

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [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> > &grid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        int m=grid.size();
	    if ( m==0 )
		    return 0;
	    int n = grid[0].size();
	    for(int i=0;i<m;i++)
		    assert(grid[i].size()==n);
	    if ( grid[0][0]== 1 )
		    return 0;

	    vector<vector<int> > path(m,vector<int>(n,0));
	    path[0][0]=1;
	    for(int i=1;i<m;i++)
	    {
		    if ( grid[i][0]==0 && path[i-1][0]==1)
			    path[i][0]=1;
		    else
			    path[i][0]=0;
	    }
	    for(int j=1;j<n;j++)
	    {
		    if (grid[0][j]==0 &&path[0][j-1]==1)
			    path[0][j]=1;
		    else
			    path[0][j]=0;
	    }
	    for(int i=1;i<m;i++)
	    {
		    for(int j=1;j<n;j++)
		    {
			    if( grid[i][j]==1 )
				    path[i][j]=0;
			    else
			    {
				    path[i][j]=path[i-1][j]+path[i][j-1];
			    }
		    }
	    }
	    return path[m-1][n-1];
    }
};

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.

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        int m=grid.size();
	    if ( m==0 )
		    return 0;
	    int n = grid[0].size();
	    for(int i=0;i<m;i++)
		    assert(grid[i].size()==n);

	    vector<vector<int> > sum(m,vector<int>(n,0));
	    sum[0][0]=grid[0][0];
        
	    for(int i=1;i<m;i++)
		    sum[i][0]=sum[i-1][0]+grid[i][0];
	    for(int i=1;i<n;i++)
		    sum[0][i]=sum[0][i-1]+grid[0][i];
	    for(int i=1;i<m;i++)
		    for(int j=1;j<n;j++)
			    sum[i][j]=min(sum[i-1][j],sum[i][j-1])+grid[i][j];
	    return sum[m-1][n-1];

    }
};