[LeetCode] 62. Unique Paths 唯一路径

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

Example 2:

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

解题思路：

Climbing Stairs二维版。计算解个数的题多半是用DP。而这两题状态也非常显然，dp[i][j]表示从起点到位置(i, j)的路径总数。DP题目定义好状态后，接下去有两个任务：找通项公式，以及确定计算的方向。

1. 由于只能向右和左走，所以对于(i, j)来说，只能从左边或上边的格子走下来：

dp[i][j] = dp[i-1][j] + dp[i][j-1]

2. 对于网格最上边和最左边，则只能从起点出发直线走到，dp[0][j] = dp[i][0] = 1

3. 计算方向从上到下，从左到右即可。可以用滚动数组实现。

 

Java Solution 1:

class Solution {
    public int uniquePaths(int m, int n) {
        if (m == 0 || n == 0) {
            return 1;
        }

        int[][] dp = new int[m][n];
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
            dp[0][i] = 1;
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }  
        }
        return dp[m - 1][n - 1];
    }
}

Java Solution 2:

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

CPP:

class Solution {
public:
    /**
     * @param n, m: positive integer (1 <= n ,m <= 100)
     * @return an integer
     */
    int uniquePaths(int m, int n) {
        // wirte your code here
        vector<vector<int> > f(m, vector<int>(n));
        
        for(int i = 0; i < n; i++)
            f[0][i] = 1;
            
        for(int i = 0; i < m; i++)
            f[i][0] = 1;
            
        for(int i = 1; i < m; i++)
            for(int j = 1; j < n; j++)
                f[i][j] = f[i-1][j] + f[i][j-1];
                
        return f[m-1][n-1];
    }
};

Python:

class Solution(object):
    def uniquePaths(self, m, n):
        dp = [[0] * n for i in xrange(m)]
        for i in xrange(m):
            for j in xrange(n):
                if i == 0 or j == 0:
                    dp[i][j] = 1
                else:
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

        return dp[m -1][n - 1]　　

Python: Time: O(m * n) Space: O(m + n)

class Solution:
    # @return an integer
    def uniquePaths(self, m, n):
        if m < n:
            return self.uniquePaths(n, m)
        ways = [1] * n

        for i in xrange(1, m):
            for j in xrange(1, n):
                ways[j] += ways[j - 1]

        return ways[n - 1]　

Python:

class Solution:
    # @return an integer
    def c(self, m, n):
        mp = {}
        for i in range(m):
            for j in range(n):
                if(i == 0 or j == 0):
                    mp[(i, j)] = 1
                else:
                    mp[(i, j)] = mp[(i - 1, j)] + mp[(i, j - 1)]
        return mp[(m - 1, n - 1)]

    def uniquePaths(self, m, n):
        return self.c(m, n)

Python: wo

class Solution(object):
    def uniquePaths(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        dp = [[0] * n for i in xrange(m)]   # 　m, n不能反了        
        for i in xrange(m):
            for j in xrange(n):                    
                if i == 0 and j == 0:
                    dp[i][j] = 1
                elif i == 0:
                    dp[i][j] = dp[i][j-1]
                elif j == 0:
                    dp[i][j] = dp[i-1][j]
                else:    
                    dp[i][j] = dp[i-1][j] + dp[i][j-1]
                    
        return dp[-1][-1]                    　

JavaScript:

/**
 * @param m: positive integer (1 <= m <= 100)
 * @param n: positive integer (1 <= n <= 100)
 * @return: An integer
 */
const uniquePaths = function (m, n) {
    var f, i, j;
    f = new Array(m);
    for (i = 0; i < m; i++) f[i] = new Array(n);
    for (i = 0; i < m; i++) {
        for (j = 0; j < n; j++) {
            if (i === 0 || j === 0) {
                f[i][j] = 1;
            } else {
                f[i][j] = f[i - 1][j] + f[i][j - 1];
            }
        }
    }
    return f[m - 1][n - 1];
}

　　　

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