LeetCode 62 — Unique Paths（C++ Java Python）

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题目：http://oj.leetcode.com/problems/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?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

题目翻译：

机器人位于m x n网格的左上角（标有'Start'）。
在任何时间，机器人只能向下或向右移动。机器人试图到达网格的右下角（标有'Finish'）。
有多少种可能的不同路径？
注意：m和n最大为100。

分析：

使用二维数组来实现。规律为除了第一行和第一列全为1外，其他格的路径数为其上一格和左一格的和。

C++实现：

<span style="font-size:14px;">    int A[m][n];    for(int i = 0; i < m; ++i)    {    A[i][0] = 1;    }    for(int i = 1; i < n; ++i)    {    A[0][i] = 1;    }    for(int i = 1; i < m; ++i)    for(int j = 1; j < n; ++j)    {    A[i][j] = A[i][j - 1] + A[i - 1][j];    }    return A[m - 1][n - 1];</span>

Java实现：

<span style="font-size:14px;">public class Solution {    public int uniquePaths(int m, int n) {        int[][] A = new int[m][n];for (int i = 0; i < m; ++i) {A[i][0] = 1;}for (int i = 1; i < n; ++i) {A[0][i] = 1;}for (int i = 1; i < m; ++i)for (int j = 1; j < n; ++j) {A[i][j] = A[i][j - 1] + A[i - 1][j];}return A[m - 1][n - 1];    }}</span>

Python实现：

<span style="font-size:14px;">class Solution:    # @return an integer    def uniquePaths(self, m, n):        paths = [[1] for i in range(m)]  # initialize the 1st column to be 1                 for i in range(n - 1):           # initialize the 1st row to be 1             paths[0].append(1)                 for i in range(m - 1):            for j in range(n - 1):                paths[i + 1].append(paths[i][j + 1] + paths[i + 1][j])                return paths[m - 1][n - 1]</span>

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