62. Unique Paths

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?

一刷
题解：
dp的经典问题，每次向右或向下走一步。第一行或者第一列走到头只有一种方法，所以初始化为1，转移方程是dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

思路一，构造m*n的矩阵保存dp[i][j]
Time complexity O(mn), space complexity O(mn)

public class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(i == 0||j==0) dp[i][j] = 1;
                else dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        
        return dp[m-1][n-1];
    }
}

思路二，由于dp[i][j] = dp[j][i]是对称的，考虑利用这个特点来save space
Time complexity O(mn), space complexity O(n)

public class Solution {
    public int uniquePaths(int m, int n) {
        if(m<0 || n<0) return 0;
        int[] dp = new int[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]; //dp[j] represent dp[i-1][j], dp[j-1] represent dp[i][j-1]
            }
        }
        
        return dp[n-1];
    }
}