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[leetcode]Unique Paths

时间:2014-11-07 11:24:26      阅读:160      评论:0      收藏:0      [点我收藏+]

标签:leetcode   算法   

问题描述:

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?

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Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.



基本思想:

本题可以这么考虑。  设f[i][j]表示从左上到i行j列的格子处的所有路径数。考虑我们可以如何到f[i][j]这个格子呢?  有两种方法:

  1. 从(i-1,j)向右走一步
  2. 从(i,j-1)向左走一步

所以我们可以列出恒等式  f[i][j] = f[i-1][j]+f[i][j-1];初始化(i,0)和(0,j)为1;如此便可以迭代的求出到最后右下时的方法数。


代码:

int uniquePaths(int m, int n) {  //c++
        if(m == 1 || n == 1)
            return 1;
        int array[m][n];
        
        //init
        for(int i = 0; i < n; i++)
            array[0][i] = 1;
        for(int i = 0; i < m; i++)
            array[i][0] = 1;
            
        for(int i = 1; i < m; i++)
            for(int j = 1; j < n; j++){
                array[i][j] = array[i-1][j]+array[i][j-1];
            }
        return array[m-1][n-1];
    }


[leetcode]Unique Paths

标签:leetcode   算法   

原文地址:http://blog.csdn.net/chenlei0630/article/details/40888961

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