地址：https://leetcode.com/problems/unique-paths/
题目：
A robot is located at the top-left corner of a 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:
- Right -> Right -> Down
- Right -> Down -> Right
- Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
理解：
用dp做就可以了。当前位置的可能等于从左边过来和从上面过来两种，则到达当前位置可能数就是dp[i-1][j]和dp[i][j-1]之和。
实现：
自己的实现，比较笨
class Solution {
public:
int uniquePaths(int m, int n) {
if (m == 1 || n == 1) return 1;
vector<vector<int>> dp(n, vector<int>(m,0));
dp[0][1] = dp[1][0] = 1;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
int left = i >= 1 ? dp[i - 1][j] : 0;
int top = j >= 1 ? dp[i][j - 1] : 0;
dp[i][j] = left + top + dp[i][j];
}
}
return dp[n - 1][m - 1];
}
};
别人的实现
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> dp(n, vector<int>(m,1));
for (int i = 1; i < n; ++i) {
for (int j = 1; j < m; ++j) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[n - 1][m - 1];
}
};
但事实上，我们在每次循环中只用到了当前行和上一行，因此可以压缩空间复杂度。
if (m > n) return uniquePaths(n, m);是为了节约空间。
class Solution {
public:
int uniquePaths(int m, int n) {
if (m > n) return uniquePaths(n, m);
vector<int> pre(m, 1), curr(m, 1);
for (int i = 1; i < n; ++i) {
for (int j = 1; j < m; ++j) {
curr[j] = curr[j - 1] + pre[j];
}
swap(curr, pre);
}
return pre[m - 1];
}
};
交换操作把curr换到了pre，而pre[0]永远都是0，后面的已经没有用了。
然而，交换操作也是多余的。因为curr是从左到右更新，因此还可以节约一个数组的空间。
class Solution {
public:
int uniquePaths(int m, int n) {
if (m > n) return uniquePaths(n, m);
vector<int> dp(m, 1);
for (int i = 1; i < n; ++i) {
for (int j = 1; j < m; ++j) {
dp[j] += dp[j - 1];
}
}
return dp[m - 1];
}
};