中山大学算法课程题目详解（第十一周）

问题描述：

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 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.

解决思路：

采用动态规划的方法，用二位数组保存到行i列j的格子的路径有多少条，则有result[i][j] = result[i-1][j] + result[i][j - 1]。而且所有的result[i][0] 以及result[0][i]都初始化为1.

具体代码实现：

int uniquePaths(int m, int n) {
int a[101][101] = { 0 };
for (int i = 0; i < m; i++) {
a[i][0] = 1;
}
for (int i = 0; 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 - 1][j] + a[i][j - 1];
}
}
return a[m - 1][n - 1];
}