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?

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

Note: m and n will be at most 100.

题意

机器人可以往右或往下走,求走到终点一共有多少条路径

分析

这道题其实可以用数学方法解决,参考:

class Solution {public:    // 一共走（m-1）+（n-1）步, 从其中选择m-1步向下    // 所以总的走法是C(m+n-2, m-1)    int uniquePaths(int m, int n) {        m--; n--;        long long mn = m + n;        long long mn_max = max(m, n);        long long mn_min = min(m, n);        long long c = 1;        for (int i = mn_max + 1; i <= mn; i++) {            c *= i;        }        for (int j = 2; j <= mn_min; j++) {            c /= j;        }        return c;    }};

但是在锻炼动态规划的能力,所以用动态规划的方法做了一遍
dp[i][j]表示走到位置i,j有多少种路径
因为只能向右向下走,所以要走到位置(i, j), 上一步可能是上方(i-1, j)和左方(i, j-1)
所以状态转移方程:
dp[i][j] = 1 , i =0&&j=0 (在第一行/第一列上走都只有一种路径)
dp[i][j] = dp[i-1][j] + dp[i][j-1] , i >=1&&j>=1

class Solution {public:    int uniquePaths(int m, int n) {        vector<vector<int>> dp(m, vector<int>(n, 1));        for (int i = 1;i < m; i++) {            for (int j = 1; j < n; j++) {                dp[i][j] = dp[i-1][j] + dp[i][j-1];            }        }        return dp[m-1][n-1];    }};