[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:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 | 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:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | 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:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 | 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:
1 2 3 4 5 6 7 8 9 10 11 | 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)
1 2 3 4 5 6 7 8 9 10 11 12 | 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:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 | 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 | 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:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 | /** * @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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