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61. Unique Paths && Unique Paths II
Unique Paths
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.
思路： 其实答案就是 C(m+n-2, m-1). 但是写程序利用动态规划会简单快捷。（给两个代码，第一个方便理解，第二个是基于第一个的优化）
1.
class Solution { // C(m+n-2, m-1)public: int uniquePaths(int m, int n) { vector<vector<int> > times(m, vector<int>(n, 0)); for(int r = 0; r < m; ++r) times[r][0] = 1; for(int c = 1; c < n; ++c) times[0][c] = 1; // 只能到 1 次 for(int r = 1; r < m; ++r) for(int c = 1; c < n; ++c) times[r][c] = times[r-1][c] + times[r][c-1]; return times[m-1][n-1]; }};2.
class Solution { // C(m+n-2, m-1)public: int uniquePaths(int m, int n) { if(m <= 0 || n <= 0) return 0; vector<int> R(n, 1); // 一行行的记录 for(int r = 1; r < m; ++r) for(int c = 1; c < n; ++c) R[c] = R[c]+ R[c-1]; return R[n-1]; }};
Unique Paths II
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.
思路：同上，只是最初初始化全 0 . 当前位置为 1 时，则当到达前位置的步数为 0.
class Solution {public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { if(!obstacleGrid.size() || !obstacleGrid[0].size()) return 0; int m = obstacleGrid.size(), n = obstacleGrid[0].size(); vector<int> R(n, 0); R[0] = 1-obstacleGrid[0][0]; for(int r = 0; r < m; ++r) for(int c = 0; c < n; ++c) { if(c > 0) R[c] = (obstacleGrid[r][c] == 1 ? 0 : (R[c] + R[c-1])); else if(obstacleGrid[r][c] == 1) R[0] = 0; } return R[n-1]; }};
61. Unique Paths && Unique Paths II