qw___的博客

62. Unique Paths 求解Start到Finish路径的总和，只允许从起点位置向下或者向右移动。 设v为路径统计矩阵，v[i][j]为经过i，j位置的路径总数。 状态转移方程：v[i][j] = v[i-1][j] + v[i][j-1];class Solution {public: int uniquePaths(int m, int n) { v

63. Unique Paths II 原地解法：class Solution {public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int m = obstacleGrid.size(); int n = obstacleGrid[0].size()