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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.
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];
}
解析
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