Leetcode题集——unique-path
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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?
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
动态规划法:定义一个二维数组 A[m][n],从左上开始依次计算每一行的值,最后返回 A[m-1][n-1]即可,递推方程是:
A[i][j]=A[i-1][j]+A[i][j-1];
int uniquePaths(int m, int n) { vector<vector<int> > v(m,vector<int>(n,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]; }
还可以继续优化,用一个长度为 n 的一维数组即可,数组元素初始值都设为1,递推方程为:A[j] += A[j-1];
也就是从第二行开始更新数组值,每次都存储当前行的值,到最后一行计算完成后,返回 A[n-1]即可。
class Solution {public: int uniquePaths(int m, int n) { vector<int> v(n,1); for(int i=1;i<m;i++) for(int j=1;j<n;j++) { v[j]=v[j]+v[j-1]; } return v[n-1]; }};
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