【LeetCode】62. Unique Paths 解題報告

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62. Unique Paths / 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?

Example 1:

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

Example 2:

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 -> Down -> Down
  2. Down -> Down -> Right
  3. Down -> Right -> Down

Example 3:

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

Example 4:

Input: m = 3, n = 3
Output: 6

Constraints:

  • 1 <= m, n <= 100
  • It’s guaranteed that the answer will be less than or equal to 2 * 10^9.

Solution: DP

思路

一開始想用 DFS，但會超時，改用 DP 紀錄路徑的數量。
能抵達格子 [m, n] 的路徑數量為 dp[m-1][n] + dp[m][n-1]，也就是上面加左邊，因為機器人只能向右或向下走。

效能

Complexity

  • Time Complexity: O(MN)
  • Space Complexity: O(MN)

LeetCode Result

Code

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