62-Unique Paths

题目

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?

分析

1. 给出一个矩阵，每一步可以选择向下或者向右
2. pathNum[i][j] = pathNum[i - 1][j] + pathNum[i][j - 1];
3. 具体可见注释

实现

/*Author:FancyDate:2017-03-20Algorithm:64-Minimum Path SumTime Complexity:*/class Solution {public:    int uniquePaths(int m, int n) {        if (m <= 0 || n <= 0)            return -1;        //保存每个位置有多少种可能性        vector<vector<int>> pathNum(m , vector<int>(n ,0));        //第一列的可能性置为1        for (int i = 0; i < m; i++)            pathNum[i][0] = 1;        //第一行的可能性置为1        for (int j = 0; j < n; j++)            pathNum[0][j] = 1;        for (int i = 1; i < m; i++)            for (int j = 1; j < n; j++)                        //每一步可以选择向下或者向右，反推就是当前位置的路径数由上位置和左位置的路径数加和                pathNum[i][j] = pathNum[i - 1][j] + pathNum[i][j - 1];          return pathNum[m - 1][n - 1];    }};