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lintcode-easy-Unique Paths

时间:2016-03-10 07:06:00      阅读:137      评论:0      收藏:0      [点我收藏+]

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A robot is located at the top-left corner of a m x ngrid (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?

 

动态规划,思路比较直接,也不用多说

public class Solution {
    /**
     * @param n, m: positive integer (1 <= n ,m <= 100)
     * @return an integer
     */
    public int uniquePaths(int m, int n) {
        // write your code here 
        
        int[][] result = new int[m][n];
        
        for(int i = 0; i < m; i++)
            result[i][0] = 1;
        for(int i = 0; i < n; i++)
            result[0][i] = 1;
        
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                result[i][j] = result[i - 1][j] + result[i][j - 1];
            }
        }
        
        return result[m - 1][n - 1];
    }
}

 

lintcode-easy-Unique Paths

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原文地址:http://www.cnblogs.com/goblinengineer/p/5260437.html

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