LeetCode | Unique Paths
2014-03-20 23:03
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题目
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.
分析
一种思路就是动态规划,另一种就是求组合数C(n+m-2, m-1),求组合数还得小心溢出。这里给出的是动态规划的解法。
代码
public class UniquePaths {
public int uniquePaths(int m, int n) {
int[] dp = new int
;
// init
dp[0] = 1;
// dp
for (int i = 0; i < m; ++i) {
for (int j = 1; j < n; ++j) {
dp[j] += dp[j - 1];
}
}
return dp[n - 1];
}
}
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.
分析
一种思路就是动态规划,另一种就是求组合数C(n+m-2, m-1),求组合数还得小心溢出。这里给出的是动态规划的解法。
代码
public class UniquePaths {
public int uniquePaths(int m, int n) {
int[] dp = new int
;
// init
dp[0] = 1;
// dp
for (int i = 0; i < m; ++i) {
for (int j = 1; j < n; ++j) {
dp[j] += dp[j - 1];
}
}
return dp[n - 1];
}
}