62. Unique Paths 动态规划算法浅谈

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.

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问题分析：

这题我一直陷入了回溯算法的做法，结果形成起来非常复杂。其实这应该算是典型的动态规划的题目，动态规划的思路在于由前面的得到现在的，关键在于写出状态转移方程。应该清醒的认识到，由于只能向下和向右走，实际上可以理解为一个单方向的走法，因此对于（i,j)格点来说，只可能由（i-1,j)和（i,j-1)得到，假设mm[i][j]表示到达格点i,j的走法，那么显然有mm[i][j]=mm[i-1][j]+mm[i][j-1];

代码如下：

class Solution {
public:
    int uniquePaths(int m, int n){
    int mm[m][n];
    for(int t=0;t         mm[t][0]=1;
    for(int t=0;t          mm[0][t]=1;
    for(int i=1;i     for(int j=1;j     mm[i][j]=mm[i-1][j]+mm[i][j-1];
    return mm[m-1][n-1];
    
}
};