leetcode_62. Unique Paths四种解法

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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 7 x 3 grid. 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

给定m*n的棋盘格，从左上角出发，只能向下或向右走，有几条路可以通向右下角

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看到一种更快速的解法，贴上来分享一下

思路：m*n的棋盘，从左上角到右下角，需要经过(m-1)个向右和(n-1)个向下，总共需要走过(m+n-2)个格子，计算从(m+n-2)个格子中选(m-1)个向右的格子，有种组合，即等价于求有多少条路径。

Class Sulotion{
    
    public int uniquePaths(int m, int n) {
        double value = 1;
        for (int i = 1; i <= n - 1; i++) {
            value *= ((double) (m + i - 1) / (double) i);
        }
        return (int) Math.round(value);
    }
}

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最开始两种方法都出现了time limit exceeded超时问题ε(┬┬﹏┬┬)3，类似图的深度优先遍历的递归算法和广度优先遍历

深度优先遍历：能往右走就走一步，能往下走就走一步，走到右下角时，返回经过进入节点到达右下角的路径数，下面附上代码，看代码还是比较好理解的

class Solution {
    public int uniquePaths(int m, int n) {
        return repeat(m, n, 0, 0);
    }
    
    private int repeat(int m, int n, int right, int down){
        if (m == right + 1 && n == down + 1)
            return 1;
        int count = 0;
        if (right+1 < m)
            count += repeat(m, n, right+1, down);
        if (down + 1 < n)
            count +=repeat(m, n, right, down+1);
        return count;
    }
}

广度优先遍历：先把左上角插入队列，当队列不空时循环以下操作，取队首，如果队首是右下角，则计数加1，如果不是右下角，则把该位置的右边和下边入队列。

import java.util.concurrent.LinkedBlockingQueue;

class Solution {
    public int uniquePaths(int m, int n) {
        if (m == 1 || n == 1)
            return 1;
        int count = 0;
        Queue queue = new LinkedBlockingQueue<>();
        int[] p = {0,0};
        queue.add(p);
        while (!queue.isEmpty()){
            int[] cur = queue.remove();
            if (cur[0] == m -1 && cur[1] == n - 1)
                count++;
            if (cur[0] < m - 1) {
                int[] next = {cur[0]+1, cur[1]};
                queue.add(next);
            }
            if (cur[1] < n - 1) {
                int[] next = {cur[0], cur[1]+1};
                queue.add(next);
            }
        }
        return count;
    }
   
}

最后一种方法，可以从左上角出发到达每个结点的路径，需要O(m*n)的空间来存放到达每个节点的路径。

动态规划：到达右下角，要么是到了右下角的上面一格，要么是到达右下角的左边一格，即到达右下角的路径数=到达右下角上面一格的路径数+到达右下角的左一格的路径数。

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