Leetcode解题笔记 62. Unique Paths [Medium] 动态规划
2017-12-06 17:02
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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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