LeetCode-062：Unique Paths

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_32360995/article/details/87249602

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 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:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

Input: m = 7, n = 3
Output: 28

题意：

在一张m*n的地图上从左上角走到右下角有几种路径？每次只能向下走或者向右走

思路：

动态规划，递推表达式：dp[i][j]=dp[i-1][j]+dp[i][j-1]~~~还可以用滚动数组来减少空间复杂度，情人节无心尝试了，明天出初试成绩，愿老天保佑我过线~~~

Code：

class Solution {
public:
    int uniquePaths(int m, int n) {
        int dp[101][101];
        memset(dp,0,sizeof(dp)); 
        for(int i=1;i<=m;i++){
            for(int j=1;j<=n;j++){
                if(i==1||j==1) dp[i][j]=1;
                else dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[m][n];
    }
};