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[Leetcode]62. Unique Paths(C++)

题目描述

题目链接:62. Unique Paths

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?

例子

例子 1

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

例子 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

例子 3

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

Constraints

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

解题思路

一般这种有多少种方法/路径的题目都可以考虑使用动态规划求解。具体做法为:

  • 使用 mxn 的二维数组表示到每一格有几种走法(初始化为 0,左上角为 1)
    • 这里实现时可以使用 (m+1)x(n+1) 的二维数组用 counts[i][j] 表示在 (i,j) (从 1 开始)的路径数
  • 从左上角到右下角遍历,更新每一个格子: counts[i][j] += (counts[i - 1][j] + counts[i][j - 1])
  • 最后返回右下角元素即可

代码如下:

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#include <vector>

class Solution {
public:
int uniquePaths(int m, int n) {
std::vector<std::vector<int>> counts(m + 1, std::vector<int>(n + 1, 0));
counts[1][1] = 1;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
counts[i][j] += (counts[i - 1][j] + counts[i][j - 1]);
}
}
return counts[m][n];
}
};
  • 时间复杂度: O(mn)
  • 空间复杂度: O(mn)

GitHub 代码同步地址: 62.UniquePaths.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions