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  • leetcode 之 Unique Paths

    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.

    思路:典型的动态规划,dp[i][j]表示到matrix[i][j]的路径个数。则dp[i][j] = dp[i-1][j] + dp[i][j-1]。

    int uniquePaths(int m, int n) {
    	if(m <=0 || n <= 0)return 0;
    	vector<vector<int> > dp(m);
    	int i,j;
    	for(i=0;i<m;i++)
    	{
    		vector<int> tmp(n,1);//至少一条
    		dp[i] = tmp;
    	}
    	for(i = 1;i < m;i++)
    	{
    		for(j = 1;j < n;j++)
    		{
    			dp[i][j] = dp[i-1][j] + dp[i][j-1];
    		}
    	}
    	return dp[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.


    思路:和上面类似,仅仅是当obstacleGrid[i][j] == 0时要把dp[i][j]=0。表示此路不通,初始化时也要注意


    class Solution {
    public:
        int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        	int rows = obstacleGrid.size();
        	if(rows <= 0)return 0;
        	int cols = obstacleGrid[0].size();
        	if(cols <= 0)return 0;
        	vector<vector<int> > dp(rows);
        	int i,j;
        	for(i = 0;i < rows;i++)
        	{
        		vector<int> tmp(cols);
        		dp[i] = tmp;
        	}
        	dp[0][0] = obstacleGrid[0][0] == 0 ? 1 : 0;
        	for(i = 1;i < rows;i++)dp[i][0] = obstacleGrid[i][0] == 0 ? dp[i-1][0] : 0;//当为0时,不能简单的初始化为1。要初始化为前面的值,由于可能被前面挡住了
        	for(j = 1;j < cols;j++)dp[0][j] = obstacleGrid[0][j] == 0 ?

    dp[0][j-1] : 0; for(i = 1;i < rows;i++) { for(j = 1;j < cols;j++) { dp[i][j] = obstacleGrid[i][j] == 0 ? dp[i-1][j] + dp[i][j-1] : 0; } } return dp[rows-1][cols-1]; } };



     

    版权声明:本文博客原创文章,博客,未经同意,不得转载。

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  • 原文地址:https://www.cnblogs.com/bhlsheji/p/4757258.html
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