[leetcode][62] 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?

Note: m and n will be at most 100.

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

Input: m = 7, n = 3
Output: 28

解析

从矩阵的左上角走到左下角有多少种走法。

参考答案

自己写的（递归搜索）：

class Solution {
    public int uniquePaths(int m, int n) {
        return backTrack(m-1, n-1, 0);
    }

    public int backTrack(int m, int n, int count) {
        int t = count;
        if (m == 0 && n == 0) {
            return t + 1;
        }

        if (m > 0) {
            t += backTrack(m-1, n, count);
        }
        if (n > 0) {
            t += backTrack(m, n-1, count);
        }

        return t;
    }
}

动态规划：

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

递归搜索会有大量重复搜索的过程，动态规划只需要遍历整个二维数组，计算出起点到每个点的路径数就能得到起点到终点的路径数。

已知条件有：

每个点的路径数等于它左边的路径数加上上边的路径数；

起点到第零列的路径数为1；

起点到第零行的路径数为1；

从这些已知条件出发就可以算出所有点的路径数，从而得到终点的路径数。