62. Unique Paths && 63. Unique Paths II

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mb5fed71eeb026e 2018-03-20 11:31:00

文章标签 i++ c++ 复杂度初始化基本语法 文章分类 C/C++ 后端开发

62. 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.

62. Unique Paths && 63. Unique Paths II_i++

解析

class Solution_62 {
public:
	int uniquePaths(int m, int n) {
		//matrix(m*n)
		vector<vector<int>> vecs(m, vector<int>(n, 1));

		for (int i = 1; i < m;i++)
		{
			for (int j = 1; j < n;j++)
			{
				vecs[i][j] = vecs[i - 1][j] + vecs[i][j - 1];
			}
		}
		return vecs[m-1][n-1];
	}

	int uniquePaths1(int m, int n) {
		vector<int > vec(n, 1); //压缩空间
		for (int i = 0; i < m; i++)
		for (int j = 0; j < n; j++)
		if (i * j != 0)
			vec[j] += vec[j - 1];
		return vec[n - 1];
	}

//      链接：https://www.nowcoder.com/questionTerminal/166eaff8439d4cd898e3ba933fbc6358
//		动态规划的复杂度也是n方，可以用排列组合的方式，复杂度为n
//		只能向右走或者向下走，所以从一共需要的步数中挑出n - 1个向下走，剩下的m - 1个就是向右走
//		其实就是从（m - 1 + n - 1）里挑选（n - 1）或者（m - 1）个，c(n, r)     n = （m - 1 + n - 1）, r = （n - 1）
//		n!/ (r!* (n - r)!)

	//注意观察到，可以发现循环的值是；C(n, m) = n!/ (m!*(n - m)!)，因为n值过大，不可以直接用公式
    //组合数学的递推公式：C(m,n)=C(m,n-1)+C(m-1,n-1)
	//C(n, 1) = n; C(n, n) = 1; C(n, 0) = 1;这样就可以用DP了

	int fun(int n, int m)
	{
		if (m==1)
		{
			return n;
		}
		if (n==m||m==0)
		{
			return 1;
		}
		return fun(n-1, m ) + fun(n - 1, m - 1);	//超时
	}
	int uniquePaths2(int m, int n) {
		
		n = (m - 1 + n - 1);
		m = (m - 1);
		
		int ret=fun(n,m);

		return ret;
	}


};

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.

解析

class Solution_63 {
public:
	int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {

		/// 使用O(n)空间的方案
		int m = obstacleGrid.size(), n = obstacleGrid[0].size();
		if (m == 0 || n == 0)
			return 0;
		vector<int> res(n, 0);
		res[0] = 1;
		for (int i = 0; i < m; i++)
		{
			for (int j = 0; j<n; j++)
			{
				if (obstacleGrid[i][j] == 1)
					res[j] = 0;
				else if (j>0)
					res[j] = res[j] + res[j - 1];
			}
		}
		return res[n - 1];

	}
};

链接：https://www.nowcoder.com/questionTerminal/3cdf08dd4e974260921b712f0a5c8752
来源：牛客网

 int uniquePathsWithObstacles(vector<vector<int> > &a) {
        int i, j, m = a.size(), n = a[0].size();
        vector<vector<int> > dp(m, vector<int>(n, 0)); // 初始化成0
        // 第一个格点的值与障碍数相反
        dp[0][0] = 1 - a[0][0];
        // 依次计算
        for(i = 0; i < m; ++i) {
            for(j = 0; j < n; ++j) {
                // 只有没有障碍才有通路
                if(a[i][j] == 0) {
                    if(i == 0 && j != 0) dp[0][j] = dp[0][j - 1]; // 左
                    else if(i != 0 && j == 0) dp[i][0] = dp[i - 1][0]; // 上
                    else if(i != 0 && j != 0) dp[i][j] += dp[i - 1][j] + dp[i][j - 1]; // 左+上
                }
            }
        }
        return dp[m - 1][n - 1];
    }