[LeetCode]--62. Unique Paths

杜鲁门

于 2016-11-06 11:39:12 发布

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分类专栏： LeetCode 文章标签： leetcode 唯一路径算法算法

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本文介绍了一个经典的算法问题：计算从网格左上角到右下角的不同路径数量。机器人只能向下或向右移动。通过动态规划的方法，我们为任意大小的网格提供了一种高效的解决方案。

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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?

这里写图片描述
Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

想法就是看每个点有多少来源，就会有多少条路径，比如[1][1]点它的来源就是[0][1]和[1][0]两点的来源之和。就大概是这么个意思。很容易理解，最后返回[m-1][n-1]这个点的值就是独立路径条数。

public class Solution {
    public int uniquePaths(int m, int n) {
        if (m == 0 || n == 0)
            return 0;
        int[][] sum = new int[m][n];
        for (int i = 0; i < m; i++)
            sum[i][0] = 1;
        for (int i = 0; i < n; i++)
            sum[0][i] = 1;
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                sum[i][j] = sum[i - 1][j] + sum[i][j - 1];
            }
        }
        return sum[m - 1][n - 1];
    }
}