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?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

思路：在只允许从左向右，从上到下移动的情况下，从mxn的矩阵（0,0）点，到（m-1，n-1）点，有多少路径。

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

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

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