62/63 Unique Paths

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 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

class Solution {public:    int uniquePaths(int m, int n) {        vector<vector<int>> dp(m, vector<int>(n, 1));    //初始化dp数组为全1        for(int i = 1; i < m; ++i) {            for(int j = 1; j < n; ++j) {                dp[i][j] = dp[i][j-1] + dp[i-1][j];        //DP状态转移方程            }        }        return dp[m-1][n-1];    }};

63 Unique Paths II

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.

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        int m = obstacleGrid.size();            //获取m,n        int n = obstacleGrid[0].size();        int x,y;        vector<vector<int>> dp(m, vector<int>(n, 1));    //初始化        for(int i = 0; i < m; ++i) {            if(obstacleGrid[i][0] == 1) {               //当边缘处出现“1”时，其后面所以格子都不可走到                for(int j = i; j < m; ++j) {                    dp[j][0] = 0;                }                break;            }        }        for(int i = 0; i < n; ++i) {                  //同上进行初始化边缘            if(obstacleGrid[0][i] == 1) {                for(int j = i; j < n; ++j) {                    dp[0][j] = 0;                }                break;            }        }        for(int i = 1; i < m; ++i) {            for(int j = 1; j < n; ++j) {                x = obstacleGrid[i][j-1] == 1? 0 : 1;    //DP状态转移方程，多加一个判断条件                y = obstacleGrid[i-1][j] == 1? 0 : 1;                dp[i][j] = dp[i][j-1]*x + dp[i-1][j]*y;            }        }        return obstacleGrid[m-1][n-1] == 1? 0 : dp[m-1][n-1];   //最后还需要查看右下角标记是否为“1”    }};

总结：DP、初始化条件、状态转移方程。