【leetcode62】 62. Unique Paths解题报告

  • 关键字:动态规划、递归
  • 难度:Medium
  • 题目大意:计算两网格之间所有可能路径,只能向右或下行走
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?


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

思路:

1、暴力递归:leetcode超时了;
2、动态规划:定义p(i, j)为 从开始位置到坐标(i,j)的所有可能路径,则:

  • i==1或j==1时,显然p(i, j) = 1;
  • 其他情况下,p(i , j) = p(i-1, j) + p(i, j-1);
    以m=7,n=3为例,dp数组可以正序填充:
DP填充.jpeg
代码实现:
解法1:
public class solution {
    public  int uniquePaths(int m, int n) {
        return process(n,m);
    }

    private  int process(int i, int j) {
        if(i==1||j==1) return 1;
        return process(i-1,j) + process(i,j-1);
    }
}

解法2:

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[n+1][m+1];
        for(int i=1;i

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