leetcode -- 62. Unique Paths

题目描述

题目难度：Medium
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 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:

• Right -> Right -> Down
• Right -> Down -> Right
• Down -> Right -> Right
• Example 2:

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

AC代码1

利用动态规划的思想，到达[x,y]坐标的方式个数等于到达[x - 1,y] 和 [x, y - 1]坐标的方式个数总和。相比其他方法下面的这种方法空间复杂度过高，这是递归方法的缺陷。

class Solution {
    int res = 0;
    public int uniquePaths(int m, int n) {
        dp(1, 1, m, n);
        return res;
    }
    
    private void dfs(int x, int y, int m, int n){
        if(x == m || y == n) {
            res++;
            return;
        }
        dp(x + 1, y, m, n);
        dp(x, y + 1, m, n);
    }
}

AC代码2

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