描述
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?
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:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right複製程式碼Example 2:
Input: m = 7, n = 3
Output: 28複製程式碼Constraints:
1 <= m, n <= 100
It's guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.複製程式碼思路
在知乎上學到了動態規劃的基本知識，就找了這道題作為練習鞏固。
此題以棋盤為dp陣列的形象描述，有助於初學者進行理解。
- 除第一行及第一列的格子只有一種到達方式外，其他的格子需要從左邊/上邊的格子讀取上一個狀態，並將這兩個狀態相加，得到最右下角的路徑數。
關係式是 dp[i][j] = dp[i-1][j] + dp[i][j-1]
class Solution {
public int uniquePaths(int m, int n) {
int dp[][] = new int[m][n];
for(int i = 0; i<n; i++){
dp[0][i] = 1;
}
for(int i = 0; i<m; i++){
dp[i][0] = 1;
}
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];
}
}複製程式碼優化
根據上一行的數值及第一列的數值（恆為1）計算第二行（下圖）
計算過程（下圖*3）
可以觀察到，第i行j列的資料生成之後，第i-1行j列的資料就可以不用儲存了。即滿足關係式dp[i] = dp[i-1] + dp[i]。照此推理過程不難推出最後一行的情況（下圖）
右下角即為所求。
class Solution {
public int uniquePaths(int m, int n) {
int dp[] = new int[n];
for(int i = 0; i<m; i++){
dp[0] = 1;
for(int j = 1; j<n; j++){
dp[j] = dp[j-1] + dp[j];
}
}
return dp[n-1];
}
}複製程式碼數學優化（來自LeetCode題目評論區）
The total number of movements of the robot is S=m+n-2, and the number of
downward movements is D=m-1. Then the problem can be seen as the number
of combinations of D positions taken from S. The solution to this
problem is C(S, D).
class Solution(object):
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
return math.factorial(m+n-2)/math.factorial(m-1)/math.factorial(n-1)複製程式碼使用數學組合方法進行計算，空間複雜度O(1)。