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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 7 x 3 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
Example 1:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
DP:二维数组class Solution {
public int uniquePaths(int m, int n) {
int[][] memo = new int[n+1][m+1];
for(int i = 1; i<= n; i++){
for(int j = 1; j<= m; j++){
memo[i][j] = 1;
}
}
for(int i = 2; i<= n; i++){
for(int j = 2; j<= m; j++){
if(i == 1 && j == 1) memo[i][j] = 1;
if((i == 1 && j != 1)) memo[i][j] = memo[i][j-1];
else if( (j == 1 && i!= 1)) memo[i][j] = memo[i-1][j];
else memo[i][j] = memo[i][j-1] + memo[i-1][j];
}
}
return memo[n][m];
}
}
DP: 优化为一维数组
class Solution {
public int uniquePaths(int m, int n) {
int[] memo = new int[m+1];
int former = 0;
for(int j = 0; j<= m; j++){
memo[j] = 1;
}
for(int i = 1; i<= n; i++){
for(int j = 1; j<= m; j++){
if((i == 1 && j != 1)) memo[j] = memo[j-1];
else if( (j == 1 && i!= 1)) former = 1;
else {
memo[j] = memo[j] + former;
former = memo[j];
}
}
}
return memo[m];
}
}
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