Unique Path

来源：互联网发布：端口号传输层编辑：程序博客网时间：2024/06/06 08:44

Unique Path是一道递归或者说动态规划的基本题

下面用c语言实现

#include <stdio.h>#include <malloc.h>#include <memory.h>int uniquePathsR(int m, int n) {if(m < 1 || n < 1) return 0;if(m == 1 || n == 1) return 1;return uniquePathsR(m - 1, n) + uniquePathsR(m, n -1);}int uniquePathsRM2(int m, int n, int *mem, int columns){if(m < 1 || n < 1) return 0;if(m == 1 || n == 1) return 1;if(mem[m * columns + n] > 0) return mem[m * columns + n];mem[m * columns + n] = uniquePathsRM2(m - 1, n, mem, columns) + uniquePathsRM2(m, n - 1, mem, columns);return mem[m * columns + n];}int uniquePathsRM(int m, int n){int *mem = (int *)malloc((m + 1) * (n + 1) * sizeof(int));memset(mem, -1, (m + 1) * (n + 1) * sizeof(int));int numOfPaths = uniquePathsRM2(m, n, mem, n);free(mem);mem = NULL;return numOfPaths;}int uniquePathsDP(int m, int n){int dp[m][n]; //变长数组，c99新特性 memset(dp, 0, sizeof(dp));for(int i = 0; i < m; i++)dp[i][0] = 1;for(int j = 0; j < n; j++)dp[0][j] = 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];}int uniquePathsDPS(int m, int n){int dp[n];for(int i = 0; i < n; i++) dp[i] = 1;for(int i = 1; i < m; i++)for(int j = 1; j < n; j++)dp[j] = dp[j] + dp[j - 1];return dp[n - 1];}int main(){printf("%d\n",uniquePathsR(3,4));printf("%d\n",uniquePathsRM(3,4));printf("%d\n",uniquePathsDP(3,4));printf("%d\n",uniquePathsDPS(3,4));return 0;}

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">用c++实现</span>

#include <iostream>#include <vector>using namespace std;int uniquePathsDP(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];}int uniquePathsDPS(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] + dp[j - 1];return dp[n - 1];}int main(){cout << uniquePathsDP(3,4) << endl;cout << uniquePathsDPS(3,4) << endl;return 0;}

直接用自顶向下的递归会超时：

Time Limit Exceeded：Last executed input:23, 12

记忆化搜索的数组用全局数组时会比较好写，如果不用全局数组则需要传递数组指针作为参数。记忆化搜索两步：取、存。

滚动数组可以优化空间。

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