Unique Paths I

　　我们来看看原题：https://leetcode.com/problems/unique-paths/。

原题

　　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?



翻译：机器人在左上角，要到达右下角，机器人每次只能向右或者向下移动，问：机器人有多少种不同的走法？

思路

　　很容易让人联想到“青蛙跳台阶”的经典问题，当然是DP（动态规划）解法。

　　子问题的划分就是——到达位置(m,n)的走法总数c(m,n)=c(m-1,n)+c(m,n-1)。

代码

　　这段代码应该可以解释他自己：

int uniquePaths(int m, int n) {
if(m < 1 || n < 1) return 0;
if(m == 1 && n == 1) return 1;

vector<vector<int>> theVector;
for(int i = 0; i < m; i++)
{
vector<int> temp(n);
theVector.push_back(temp);
}

for(int i = 0; i < n; i++)
theVector[0][i] = 1;
for(int i = 0; i < m; i++)
theVector[i][0] = 1;

for(int i = 1; i < m; i++)
for(int j = 1; j < n; j++)
theVector[i][j] = theVector[i - 1][j] + theVector[i][j - 1];

return theVector[m - 1][n - 1];
}

　　接下来看看问题2去~