62. Unique Paths

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每日一题 2019 - 04 - 05

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?

img
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:

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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:

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

解法：

这是一道基础的动态规划类的问题，需要找出对应的等式递推关系就可以解决问题；

思路：

  • 到达[i,j]位置的 robot 必然是从 [i,j-1] 或者 [i-1,j] 位置到来，那么证明到这个位置可以有 dp[i-1,j] + [i,j-1] 种走法

所以从上可知：递推公式为 dp[i,j] = dp[i-1,j] + dp[i,j-1]

代码：

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class Solution {
public:
    int uniquePaths(int m, int n) {
        int i,j;
        int dp[m+1][n+1];    //表示 i 行 j 列的不同路径
        for (i = 1,j = 1;j < n + 1;j++) {
            dp[i][j] = 1;
        }
        for (i = 1,j = 1;i < m + 1;i++) {
            dp[i][j] = 1;
        }
        for (i = 2;i < m + 1;i++)
        {
            for (j = 2;j < n + 1;j++)
            {
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m][n];
    }
};
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