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
1 class Solution {2 public:3 int uniquePaths(int m, int n) {4 if(m == 1 || n == 1)return 1;5 else return uniquePaths(m, n - 1) + uniquePaths(m - 1, n);6 }7 }; 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
1 class Solution { 2 public: 3 int uniquePaths(int m, int n) { 4 vector > dp(m+1, vector (n+1, 1)); 5 for(int i = 2; i <= m; i++) 6 for(int j = 2; j <= n; j++) 7 dp[i][j] = dp[i-1][j] + dp[i][j-1]; 8 return dp[m][n]; 9 }10 }; 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:
1 class Solution { 2 public: 3 int uniquePaths(int m, int n) { 4 vector dp(n+1, 1); 5 for(int i = 2; i <= m; i++) 6 for(int j = 2; j <= n; j++) 7 dp[j] = dp[j] + dp[j-1]; 8 return dp[n]; 9 }10 }; 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
1 class Solution { 2 public: 3 int uniquePaths(int m, int n) { 4 return combination(m+n-2, m-1); 5 } 6 7 int combination(int a, int b) 8 { 9 if(b > (a >> 1))b = a - b;10 long long res = 1;11 for(int i = 1; i <= b; i++)12 res = res * (a - i + 1) / i;13 return res;14 }15 };
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
1 class Solution { 2 public: 3 int uniquePathsWithObstacles(vector > &obstacleGrid) { 4 int m = obstacleGrid.size(), n = obstacleGrid[0].size(); 5 vector dp(n+1, 0); 6 dp[1] = (obstacleGrid[0][0] == 0) ? 1 : 0; 7 for(int i = 1; i <= m; i++) 8 for(int j = 1; j <= n; j++) 9 if(obstacleGrid[i-1][j-1] == 0)10 dp[j] = dp[j] + dp[j-1];11 else dp[j] = 0;12 return dp[n];13 }14 };
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