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 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
算法1：最容易想到的是递归解法，uniquePaths(m, n) = uniquePaths(m, n-1) + uniquePaths(m-1, n), 递归结束条件是m或n等于1，这个方法oj超时了
class Solution {
public:
int uniquePaths(int m, int n) {
if(m == || n == )return ;
else return uniquePaths(m, n - ) + uniquePaths(m - , n);
}
};
算法2：动态规划，算法1的递归解法中，其实我们计算了很多重复的子问题，比如计算uniquePaths(4, 5) 和 uniquePaths(5, 3)时都要计算子问题uniquePaths(3, 2)，再者由于uniquePaths(m, n) = uniquePaths(n, m)，这也使得许多子问题被重复计算了。要保存子问题的状态，这样很自然的就想到了动态规划方法，设dp[i][j] = uniquePaths(i, j)， 那么动态规划方程为：
- dp[i][j] = dp[i-1][j] + dp[i][j-1]
- 边界条件：dp[i][1] = 1, dp[1][j] = 1
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int> > dp(m+, vector<int>(n+, ));
for(int i = ; i <= m; i++)
for(int j = ; j <= n; j++)
dp[i][j] = dp[i-][j] + dp[i][j-];
return dp[m][n];
}
};
上述过程其实是从左上角开始，逐行计算到达每个格子的路线数目，由递推公式可以看出，到达当前格子的路线数目和两个格子有关：1、上一行同列格子的路线数目；2、同一行上一列格子的路线数目。据此我们可以优化上面动态规划方法的空间:
class Solution {
public:
int uniquePaths(int m, int n) {
vector<int>dp(n+, );
for(int i = ; i <= m; i++)
for(int j = ; j <= n; j++)
dp[j] = dp[j] + dp[j-];
return dp[n];
}
};
算法3：其实这个和组合数有关，对于m*n的网格，从左上角走到右下角，总共需要走m+n-2步，其中必定有m-1步是朝右走，n-1步是朝下走，那么这个问题的答案就是组合数：
, 这里需要注意的是求组合数时防止乘法溢出 本文地址
class Solution {
public:
int uniquePaths(int m, int n) {
return combination(m+n-, m-);
}
int combination(int a, int b)
{
if(b > (a >> ))b = a - b;
long long res = ;
for(int i = ; i <= b; i++)
res = res * (a - i + ) / i;
return res;
}
};
Follow up for "Unique Paths":
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1 and 0 respectively in the grid.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is 2.
Note: m and n will be at most 100.
这一题可以完全采用和上一题一样的解法，只是需要注意dp的初始化值，和循环的起始值
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m = obstacleGrid.size(), n = obstacleGrid[].size();
vector<int>dp(n+, );
dp[] = (obstacleGrid[][] == ) ? : ;
for(int i = ; i <= m; i++)
for(int j = ; j <= n; j++)
if(obstacleGrid[i-][j-] == )
dp[j] = dp[j] + dp[j-];
else dp[j] = ;
return dp[n];
}
};
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