62. Unique Paths
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:
- Right -> Right -> Down
- Right -> Down -> Right
- Down -> Right -> Right
Example 2:
Input: m = 7, n = 3 Output: 28
思路：
题目意思是从一个m x n 的矩阵的左上方走到右下方 ，只能向下或者向右走，一共有多少种走法。同样，先看这个问题的最后一步，最后一步是走到第m行n列这个格子，问有多少种走法，那这个格子只能是从第m-1行第n列往下走一步或者第m行第n-1列往右走一步，由加法原则就可以知道，到达m行n列是这两种走法的和。所以就变成了得知道m行n-1列和m-1列n行的走法，这就是子问题，状态转移方程就是
dp[i][j]=dp[i-1][j] dp[i][j-1]，dp[i][j]代表了走到第i行第j列的走法。然后考虑初始条件和边界条件，dp[0][j]和dp[i][0]不能通过状态转移方程求出，所以这两个就是初始条件。 （可以在维度上优化为一维，状态转移方程变成dp[i] = dp[i-1] dp[i]，从左往右扫或者从上往下扫。以从上往下扫为例，当前dp[i]代表的就变成了走到dp[i]需要从左边往右走一步，也就时dp[i-1]，和上一层的第i列那个格子往下走一步，也就是上一个dp[i]）
代码：
go：
代码语言：javascript复制/*func uniquePaths(m int, n int) int {
if m <= 0 || n <= 0 {return 0}
// initial dp array
dp := make([][]int, m)
for i := 0; i < len(dp); i {
dp[i] = make([]int, n)
}
// find unique path
for i := 0; i < m; i {
for j := 0; j < n; j {
if i == 0 || j == 0 { // initialization
dp[i][j] = 1
} else {
dp[i][j] = dp[i][j-1] dp[i-1][j]
}
}
}
return dp[m-1][n-1]
}*/
func uniquePaths(m int, n int) int {
if m <= 0 || n <= 0 {
return 0
}
dp := make([]int, n)
// find unique path
for i := 0; i < m; i {
for j := 0; j < n; j {
if j == 0 { // initialization
dp[j] = 1
} else {
dp[j] = dp[j] dp[j-1]
}
}
}
return dp[n-1]
}```
