Leetcode62-63 Unique Paths

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.

分析

这是一个关于动态规划的计算路径的问题，一个机器人只可以向下和向右行走，问最终到达终点有多少条路径。
可以观察到，对于每一个格子[i, j]，从起始位置即[0, 0]到达该点的路径数等于到达[i - 1][j]的路径数加上到达[i][j - 1]的路径数。注意边界情况即可。
因此，定义result[i][j]表示从起始点到当前点的路径数，可以写出状态转移方程如下：

result[0][j] = 1;       // j < nresult[i][0] = 1;       // i < mresult[i][j] = result[i - 1][j] + result[i][j - 1];     // 0 < i < m && 0 < j < n

最终的结果即是result[m - 1][n - 1]

代码：

class Solution {public:    int uniquePaths(int m, int n) {        if (m == 0 || n == 0) return 0;        vector<int> temp(n, 0);        vector<vector<int>> result(m, temp);        for (int i = 0; i < m; i++) {            for (int j = 0; j < n; j++) {                if (i == 0 && j == 0) {                    result[i][j] = 1;                    continue;                }                int leftRoute = 0, topRoute = 0;                if (i - 1 >= 0) {                    leftRoute = result[i - 1][j];                }                if (j - 1 >= 0) {                    topRoute = result[i][j - 1];                }                result[i][j] = leftRoute + topRoute;            }        }        return result[m - 1][n - 1];    }};

改进

因为每一个各自的路径数只依赖与result[i - 1][j] 和 result[i][j - 1]。因此，可以只用一维数组来实现状态的转移。
代码如下：

class Solution {public:    int uniquePaths(int m, int n) {        if (m == 0 || n == 0) return 0;        vector<int> result(n, 1);         for (int i = 1; i < m; i++) {            for (int j = 1; j < n; j++) {                result[j] = result[j] + result[j - 1];            }        }        return result[n - 1];    }};

运行结果

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

题目是从第一道题继承而来，只是，对题目的解增加了限制条件，即引入了障碍，如果一个格子中有障碍物，那么这个格子不可达。
延续同一样的思路，略加变化，可以写出状态转移方程如下：

result[i][0] = result[i - 1][0]     // i < nresult[0][j] = result[0][j - 1]     // j < mresult[i][j] = obstacleGrid[i][j]? 0 : result[i - 1][j] + result[i][j - 1]  // i > 0 && j > 0

代码：

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

改进

改进的思路与上一题相同，是对于空间复杂度上的改进，可以只用一维数组来记录路径的信息。因为每一个格子只依赖于[i - 1][j] 和 [i][j - 1]。
代码如下：

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

运行结果