Leetcode 62. Unique Paths
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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.
s思路:
1. dp做。因为到终点终点只有两种方式:上边和左边,所以到终点的数量=上边点的数量+左边点的数量。这个过程是recursive的。需要注意的是:有可能某个点的数量需要多次计算,为了减少运算,可以用memorization,这就是所谓的bottom-up DP.
2. 当然,好理解的DP还是top-down,即:计算每个点的次数,先依次计算和起点同一行、同一列的所有点的次数,然后后面的位置都可以由:dp[i][j]=dp[i-1][j]+dp[i][j-1]得到。
3. dp空间上可以优化。因为每个数之和左边和上边数有关系,即:每个数只被使用一次,这种情况我们可以把2d的dp修改成1d的dp。见方法2。
//方法1: 2d top-down DPclass Solution {public: int uniquePaths(int m, int n) { // int dp[m][n]={0}; for(int i=0;i<n;i++) dp[0][i]=1; for(int i=1;i<m;i++) dp[i][0]=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]; }};//方法2: 1d top-down DPclass Solution {public: int uniquePaths(int m, int n) { // int dp[n+1]={0}; dp[1]=1; for(int i=0;i<m;i++){ for(int j=2;j<=n;j++) dp[j]=dp[j]+dp[j-1]; } return dp[n]; }};
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