[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];
}
