# Unique Paths - tags: [[DP_Matrix](# "根据动态规划解题的四要素，矩阵类动态规划问题通常可用 f[x][y] 表示从起点走到坐标(x,y)的值")] ### Source - lintcode: [(114) Unique Paths](http://www.lintcode.com/en/problem/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? Note m and n will be at most 100. ~~~ ### 题解 题目要求：给定*m x n*矩阵，求左上角到右下角的路径总数，每次只能向左或者向右前进。按照动态规划中矩阵类问题的通用方法： 1. State: f[m][n] 从起点到坐标(m,n)的路径数目 1. Function: f[m][n] = f[m-1][n] + f[m][n-1] 分析终点与左边及右边节点的路径数，发现从左边或者右边到达终点的路径一定不会重合，相加即为唯一的路径总数 1. Initialization: f[i][j] = 1, 到矩阵中任一节点均至少有一条路径，其实关键之处在于给第0行和第0列初始化，免去了单独遍历第0行和第0列进行初始化 1. Answer: f[m - 1][n - 1] ### C++ ~~~ class Solution { public: /** * @param n, m: positive integer (1 <= n ,m <= 100) * @return an integer */ int uniquePaths(int m, int n) { if (m < 1 || n < 1) { return 0; } vector<vector<int> > ret(m, vector<int>(n, 1)); for (int i = 1; i != m; ++i) { for (int j = 1; j != n; ++j) { ret[i][j] = ret[i - 1][j] + ret[i][j - 1]; } } return ret[m - 1][n - 1]; } }; ~~~ ### 源码分析 1. 异常处理，虽然题目有保证为正整数，但还是判断一下以防万一 1. 初始化二维矩阵，值均为1 1. 按照转移矩阵函数进行累加 1. 任何`ret[m - 1][n - 1]`