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?

不同路径 - 图1

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:

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

Example 2:

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

题目大意

一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。问总共有多少条不同的路径？

解题思路

  • 这是一道简单的 DP 题。输出地图上从左上角走到右下角的走法数。
  • 由于机器人只能向右走和向下走，所以地图的第一行和第一列的走法数都是 1，地图中任意一点的走法数是 dp[i][j] = dp[i-1][j] + dp[i][j-1]

代码解析

class Solution {

    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        //初始状态
        for (int l = 0; l < m; l++) {
            dp[l][0] = 1;
        }
        for (int r = 1; r < n; r++) {
            dp[0][r] = 1;
        }

        //状态转移方程
        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];
    }
}