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?

到达某一格的路径数量等于它的上面和左边的路径数之和，结束条件是走到行或者列的边缘。

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

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：dp[i] +=dp[i-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
dp[n-1]即可。

public int uniquePaths(int m, int n) {
if(m==0 || n==0) return 0;
if(m==1 || n==1) return 1;
int[] dp = new int
;
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-1];
}
return dp[n-1];
}