【LeetCode】动态规划：第62题. Unique Paths

62. 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?

大意就是一个只能往下或右边走的机器人在m x n的矩形中的最左上角，想要移动到最右下角，求他的唯一的路径的个数是多少。

思路：

因为这个机器人只能往下或右边走，那么当他到达某一点的时候，他肯定是来自于上面或左边的那个点。如果dp[i][j]可以用来表示在(i,j)这个点的唯一路径个数，也就是说状态方程就是dp[i][j] = dp[i][j - 1] + dp[i - 1][j]

实现：

    public int uniquePaths(int m, int n) {
     
        Integer[][] map = new Integer[m][n];
        //初始化
        for(int i = 0; i<m;i++){
     
            map[i][0] = 1;
        }
        for(int j= 0;j<n;j++){
     
            map[0][j]=1;
        }
        //DP
        for(int i = 1;i<m;i++){
     
            for(int j = 1;j<n;j++){
     
                map[i][j] = map[i-1][j]+map[i][j-1];
            }
        }
        return map[m-1][n-1];
    }