Problem
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?
Note
简单的动规题目,建立dp数组。dp[i][j]
坐标为矩阵的坐标,值为从左上角到这一格的走法总数。赋初值,最上一行和最左列的所有格子的走法都只有一种,其余格子的走法等于其左边格子走法与上方格子走法之和。最后,返回dp[m-1][n-1]
即可。
Solution
二维DP
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];
}
}
一维DP
public class Solution {
public int uniquePaths(int m, int n) {
int[] dp = new int[n];
Arrays.fill(dp, 1);
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[j] += dp[j-1];
}
}
return dp[n-1];
}
}
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