第十周

62. Unique Paths

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Description Hints Submissions Solutions

• Total Accepted: 132181
• Total Submissions: 328812
• Difficulty: Medium
• Contributor: LeetCode

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.

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思路：

建立一个m×n、值均为1的数组，从第二行、第二列开始，每个位置的元素的值为该元素的上方和左方的值之和，这样即是将向下行或向右行的路径都记录到所走到的位置上。

代码：

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector> p(m, vector(n, 1));
        for(int i = 1; i < m; i ++){
            for(int j = 1; j < n; j ++){
                p[i][j] = p[i - 1][j] + p[i][j - 1];
            }
        }
        return p[m - 1][n - 1];
        
    }
    
};

63. Unique Paths II

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Description Hints Submissions Solutions

• Total Accepted: 98066
• Total Submissions: 313014
• Difficulty: Medium
• Contributor: LeetCode

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.

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思路：

建立一个（m+1）×（n+1）、值均为0的数组p，其中p[0][1] = 1，根据数组obstacleGrid行为i、列为j的值，若为0（即可通行）则赋值p[i][j]为1，若为1则赋值为0。再从数组p的第二行、第二列开始，若p[i][j]的值为1，则将该位置的值赋为p[i-1][j]和p[i][j-1]的和。

代码：

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