leetcode_c++：Unique Paths（062）

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?

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算法

DP
复杂度：O（nm）

1. 机器人要到(i,j)点，可以选择(i-1,j)和(i,j-1)
   到(i,j)的唯一路径数
   dp[i][j] = dp[i-1][j] + dp[i][j-1]
   dp[i][j] 是从(0,0) 到（i，j）的唯一路径

2. 网格最上边和最左边，则只能从起点走直线，dp[0][j]=dp[i][0]=1

3. 计算方向从上到下，从左到右，滚动数组

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class Solution {public:    int uniquePaths(int m, int n) {        int dp[m][n];        // 初始化dp，m*1全是1的情况        for(int i=0;i<m;i++)            dp[i][0]=1;        // 初始化dp,1*n全是1的情况        for(int j=0;j<n;j++){            dp[0][j]=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];    }};