Unique Paths I
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.
解题思路：
Climbing Stairs二维版。计算解个数的题多半是用DP。而这两题状态也非常显然，dp[i][j]表示从起点到位置(i, j)的路径总数。DP题目定义好状态后，接下去有两个任务：找通项公式，以及确定计算的方向。
1. 由于只能向右和左走，所以对于(i, j)来说，只能从左边或上边的格子走下来：
dp[i][j] = dp[i-1][j] + dp[i][j-1]
2. 对于网格最上边和最左边，则只能从起点出发直线走到，dp[0][j] = dp[i][0] = 1
3. 计算方向从上到下，从左到右即可。可以用滚动数组实现。
Java Solution 1:
class Solution {
public int uniquePaths(int m, int n) {
if (m == 0 || n == 0) {
return 1;
}
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
dp[i][0] = 1;
}
for (int i = 0; i < n; i++) {
dp[0][i] = 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];
}
public static void main(String args[]) {
Solution sol = new Solution();
System.out.println(sol.uniquePaths(3, 3));
}
}
Java Solution 2:
class Solution {
public int uniquePaths(int m, int n) {
int[][] dp = new int[m][n];
int i, j;
for (i = 0; i < m; ++i) {
for (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];
}
}