62. Unique Paths&&63. Unique Paths II 路径-优快云博客

本文介绍了一个经典的动态规划问题Unique Paths及其变种Unique Paths II的解决方案。通过分析网格中从起点到终点的不同路径数量，特别是考虑到障碍物的情况。提供了一种高效的算法实现，包括初始化边界条件的方法及如何利用递归关系更新状态矩阵。

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Unique Paths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.

Unique Paths II题目：

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

分析：

这两道都比较简单，对于动态规划题应该细心寻找dp[i][j]间的递归关系，本题就是dp[i][j]=dp[i-1][j]+dp[i][j-1]
有障碍时直接让dp[i][j]等于0即可

代码：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m=obstacleGrid.size();
        int n=obstacleGrid[0].size();
        vector<vector<int>> dp(m,vector<int>(n,0));
        dp[0][0]=1;
        for(int i=0;i<m;i++){
            dp[i][0]=1;
            if(obstacleGrid[i][0]==1){
                dp[i][0]=0;
                break;
            }
        }
        for(int i=0;i<n;i++){
            dp[0][i]=1;
            if(obstacleGrid[0][i]==1){
                dp[0][i]=0;
                break;
            }
        }
        for(int i=1;i<m;i++){
            for(int j=1;j<n;j++){
                if(obstacleGrid[i][j]==1)
                    dp[i][j]=0;
                else
                    dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
};

看别人的代码时，发现了更为简便的写法：

time is O(nm) , space is O(nm) .

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        int m = obstacleGrid.size() , n = obstacleGrid[0].size();
        vector<vector<int>> dp(m+1,vector<int>(n+1,0));
        dp[0][1] = 1;
        for(int i = 1 ; i <= m ; ++i)
            for(int j = 1 ; j <= n ; ++j)
                if(!obstacleGrid[i-1][j-1])
                    dp[i][j] = dp[i-1][j]+dp[i][j-1];
        return dp[m][n];
    }
};