Leetcode解题笔记 62. Unique Paths [Medium] 动态规划
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解题思路
设v[i][j] 为到达第i,j格的独立路径数,则v[i][j] = v[i][j - 1] + v[i - 1][j],原因是只会有从上面一格往下走或者从右面一格往左走两种情况
代码
class Solution {public: int uniquePaths(int m, int n) { vector<vector<int>> v(m + 1); for (int i = 0; i < m + 1; i++) { v[i].resize(n + 1); } for (int i = 0; i <= m; i++) { v[i][0] = 1; } for (int i = 0; i <= n; i++) { v[0][i] = 1; } for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j ++) { v[i][j] = v[i - 1][j] + v[i][j - 1]; } } return v[m-1][n-1]; }};
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