leetCode练习（62）

题目：Unique Paths

难度：medium

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.

解题思路：

       一开始正向思维，只想到用dfs穷举，但是这样的话也太太多了，当初只有n的时候，N皇后问题就已经超慢了，更何况现在是N*N.

       后来想到逆向思考，最后一行和最后一列到达终点的路径都是0。同时想到，一个点b[I][j]到终点的路径=b[I+1][j]+b[I][j+1]。发现了动态规划的方法。

      具体代码如下：

public static int uniquePaths(int m, int n) {        int[][] board=new int[m][n];        int i,j;        if(m==1||n==1){        return 1;        }        //棋盘最下一行 最右一列都只有一条路        for(i=0;i<m;i++){        board[i][n-1]=1;        }        for(i=0;i<n;i++){        board[m-1][i]=1;        }        //从m-2行开始        for(i=m-2;i>=0;i--){        for(j=n-2;j>=0;j--){        board[i][j]=board[i+1][j]+board[i][j+1];        }        }        return board[0][0];    }