Unique Paths I
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我们来看看原题:https://leetcode.com/problems/unique-paths/。
原题
A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).
How many possible unique paths are there?
翻译:机器人在左上角,要到达右下角,机器人每次只能向右或者向下移动,问:机器人有多少种不同的走法?
思路
很容易让人联想到“青蛙跳台阶”的经典问题,当然是DP(动态规划)解法。
子问题的划分就是——到达位置(m,n)的走法总数c(m,n)=c(m-1,n)+c(m,n-1)。
代码
这段代码应该可以解释他自己:
int uniquePaths(int m, int n) { if(m < 1 || n < 1) return 0; if(m == 1 && n == 1) return 1; vector<vector<int>> theVector; for(int i = 0; i < m; i++) { vector<int> temp(n); theVector.push_back(temp); } for(int i = 0; i < n; i++) theVector[0][i] = 1; for(int i = 0; i < m; i++) theVector[i][0] = 1; for(int i = 1; i < m; i++) for(int j = 1; j < n; j++) theVector[i][j] = theVector[i - 1][j] + theVector[i][j - 1]; return theVector[m - 1][n - 1]; }
接下来看看问题2去~
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