62. Unique Paths \ 63. Unique Paths II

62. Unique Paths

题目描述

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?

一个矩阵mxn，从左上角到右下角有多少条路径。

解法描述

法一：使用深度搜索。这种方法会超时。

class Solution {public:    void uniquePathsNum(int &res, int m, int n, int i, int j, int stt) {        if(i > m || j > n) return;        if(stt >=  m+n - 2) {            if(m == i && n == j) {res++; return;}            else return;        }            uniquePathsNum(res, m, n, i+1, j, stt+1);        uniquePathsNum(res, m, n, i, j+1, stt+1);    }    int uniquePaths(int m, int n) {        int res = 0;        uniquePathsNum(res, m, n, 1, 1, 0);        return res;    }};

法二：使用DP来计算。

class Solution {public:    int uniquePaths(int m, int n) {        /*        int **tmp = new int*[m+1];        for(int k = 0; k <= m; k++)            tmp[k] = new int[n+1];        */            int tmp[101][101] = {0};        tmp[1][1] = 1;        // cout << tmp[0][1] << endl;        for(int i = 1; i <= m; i++) {            for(int j = 1; j <= n; j++) {                tmp[i][j] += (tmp[i-1][j] + tmp[i][j-1]);               // cout << i << "  " << j << "   "  << tmp[i][j] << "  "  << tmp[i-1][j] << "   "  << tmp[i][j-1] << endl;             }        }        int res = tmp[m][n];        /*        for(int i = 0; i <= m; i++)            delete tmp[i];        delete tmp;        */        return res;    }};

使用空间复杂度O(n)的代码：

class Solution {public:    int uniquePaths(int m, int n) {        int *memo = new int[n];        for(int i = 0; i < n; i++)            memo[i] = 1;        for(int i = 1 ; i < m; i++)            for(int j = 1; j < n; j++)                memo[j] += memo[j-1];        return memo[n-1];    }};

当然这道题还有方法，那就是使用高中学过的排列组合。

class Solution {public:    int uniquePaths(int m, int n) {        if(m == 1 || n == 1)            return 1;        m--;        n--;        if(m < n) {              // Swap, so that m is the bigger number            m = m + n;            n = m - n;            m = m - n;        }        long res = 1;        int j = 1;        for(int i = m+1; i <= m+n; i++, j++){       // Instead of taking factorial, keep on multiply & divide            res *= i;            res /= j;        }        return (int)res;    }};

class Solution {public:    int uniquePaths(int m, int n) {        return nCr (m + n - 2, min(m, n) - 1);    }private:    int nCr (int n, int r) {        long long_result = 1;        for (int i = 0; i != r; ++i) {            // from n - r + 1 (when i = 0) to n (when i = r - 1)            long_result *= (n - r + 1 + i);            // from 1 (when i = 0)         to r (when i = r - 1)            long_result /= (i + 1);        }        return (int) long_result;    }};

63. Unique Paths II

题目描述

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.

实现代码

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& g) {        int row = g.size();        int column = row > 0?g[0].size():0;        int v[101][101] = {0};        v[1][1] = 1 - g[0][0];        for(int i = 1; i <= row; i++) {            for(int j = 1; j <= column; j++) {                if((i != 1 || j != 1) && g[i-1][j-1]) v[i][j] = 0;                else v[i][j] += v[i-1][j] + v[i][j-1];             }        }        return v[row][column];    }};