LeetCode 62. Unique Paths

Solution1：我的未能AC的答案
递归超时了！！！

class Solution {
public:int uniquePaths(int m, int n) {int res = 0;my_unique(n, m, res, 0, 0);return res;}void my_unique(int row, int col, int& res, int i, int j) {if (i >= row || j >= col)return;else if (i == row - 1 && j == col - 1) {res = res + 1;return;}else {my_unique(row, col, res, i + 1, j);my_unique(row, col, res, i, j + 1);}}
};

Solution2：
参考网址：http://www.cnblogs.com/grandyang/p/4353555.html
利用排列组合的公式：

Amn=n!(n−m)!Anm=n!(n−m)!

A_n^m=\frac{n!}{(n-m)!}

Cmn=Amnm!=n!m!(n−m)!=n(n−1)...(n−m+1)m!Cnm=Anmm!=n!m!(n−m)!=n(n−1)...(n−m+1)m!

C_n^m=\frac{A_n^m}{m!}=\frac{n!}{m!(n-m)!}=\frac{n(n-1)...(n-m+1)}{m!}
机器人一共走了m+n-2步，任意挑m-1步向右即可！
CmnCnmC_n^m和 Cn−mnCnn−mC_n^{n-m}是一样的，但为了防止溢出在循环时利用较小值进行计算!!!

class Solution {
public:int uniquePaths(int m, int n) {double a = 1, b = 1, N = m + n - 2, M = m > n? n - 1: m - 1;for (int i = 1; i <= M; i++) {a *= N + 1 - i;b *= i;}return a / b;}
};

Solution3:
动态规划，参考网址：http://www.cnblogs.com/grandyang/p/4353555.html

//DP维护二维数组
class Solution {
public:int uniquePaths(int m, int n) {vector<vector<int> > dp(m, vector<int> (n, 1));for (int i = 1; i < m; ++i) for (int j = 1; j < n; ++j) dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; return dp[m - 1][n - 1];}
};//DP维护一维数组
class Solution {
public:int uniquePaths(int m, int n) {vector<int> dp(n, 1);for (int i = 1; i < m; ++i) {for (int j = 1; j < n; ++j) {dp[j] += dp[j - 1]; }}return dp[n - 1];}
};

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【动态规划】LeetCode 62. Unique Paths

LeetCode 62. Unique Paths

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