【Leetcode】62. Unique Paths(动态规划)

谁借莪１个温暖的怀抱￠ 2022-03-01 08:48 160阅读 0赞

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?

![robot_maze.png][]  
Above is a 7 x 3 grid. How many possible unique paths are there?

**Note:** *m* and *n* will be at most 100.

**Example 1:**

Input: m = 3, n = 2
    Output: 3
    Explanation:
    From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
    1. Right -> Right -> Down
    2. Right -> Down -> Right
    3. Down -> Right -> Right

**Example 2:**

Input: m = 7, n = 3
    Output: 28

题目大意：

机器人希望从矩阵的左上角走到左下角，并且机器人只有两种行走方式，即向下和向右，问有多少种走法。

解题思路：

本题是典型的动态规划，即当前结果与其相邻的位置的结果存在联系。我们可知本位置x(i,j)的结果是邻近的两个位置x1(i-1,j),x2(i,j-1)位置的方式总和。因为机器人如果想走到x(i,j)就必须要经过x1(i-1,j),x2(i,j-1)。所以我们在开始初始化第一行和第一列为1就可以求出矩阵中全部位置的结果。

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

[robot_maze.png]: https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png