解题思路

题目给出了一个重要条件“机器人每次只能向下或者向右移动一步”由这个条件可以得出

  1. 当前的路径和与他的左边和上边的路径有关。[i - 1][j] 和 [i][j - 1]
  2. dp使用二维数组,dp[i][j]能到此处的所有路径和
  3. 得出推导公式为: dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
  4. 遍历顺序：顺序遍历
  5. 初始化：dp[0][0] = 1

代码

Python

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class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        box = [[0] * n for _ in range(m)]
        box[0][0] = 1
        for i in range(m):
            for j in range(n):
                if i == j == 0: continue
                if 0 <= i - 1: box[i][j] += box[i - 1][j]
                if 0 <= j - 1: box[i][j] += box[i][j - 1]
        return box[m - 1][n - 1]

C++

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

Go

func uniquePaths(m int, n int) int {
    dp := make([][]int, m)
    for i := range dp {
        dp[i] = make([]int, n)
    }
    dp[0][0] = 1
    for i := 0; i < m; i++{
        for j := 0; j < n; j++ {
            if j == 0 && i == 0 {continue}
            if 0 <= i - 1 {dp[i][j] += dp[i - 1][j]}
            if 0 <= j - 1 {dp[i][j] += dp[i][j - 1]}
        }
    }
    return dp[m - 1][n - 1]
}