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?
其实对每个格子,到达它要么从它上面到,要么从左边到,这是一个动态规划问题,
f[i][j] = f[i-1][j]+f[i][j-1];
f[1][1:n] = f[1:m]f[1] = 1;
class Solution {
public:
int uniquePaths(int m, int n) {
int M[101][101];
for (int i = 1; i <= m; i++) {
M[i][1] = 1;
}
for (int i = 1; i <= n; i++) {
M[1][i] = 1;
}
for (int i = 2; i <= m; i++) {
for (int j = 2; j <= n; j++) {
M[i][j] = M[i-1][j]+M[i][j-1];
}
}
return M[m][n];
}
};
优化降低空间复杂度:
class Solution {
public:
int uniquePaths(int m, int n) {
int M[101];
for (int i = 1; i <= n; i++) {
M[i] = 1;
}
for (int i = 2; i <= m; i++) {
for (int j = 2; j <= n; j++) {
M[j] = M[j]+M[j-1];
}
}
return M[n];
}
};
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.
现在我们为地图增加路障,其实算法思想仍然没有改变。
f[i][j] = f[i-1][j]+f[i][j-1];
但是初始化的第一行和第一列时”
如果obstacleGrid[0][k] = 1, 则 f[0][k:n] = 0;
如果obstacleGrid[k][0] = 1, 则 f[k:m][0] = 0;
且对每个obstacleGrid[i][j] = 1 -> f[i][j] = 0;
空间复杂度优化后代码如下
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
if(m < 1) return 0;
int n = obstacleGrid[0].size();
int M[101];
int k = 1;
for (int i = 0; i < n; i++) {
if (obstacleGrid[0][i]) {
k = 0;
}
M[i] = k;
}
k = M[0];
for (int i = 1; i < m; i++) {
if (obstacleGrid[i][0]) {
k = 0;
}
M[0] = k;
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j]) {
M[j] = 0;
} else {
M[j] = M[j]+M[j-1];
}
}
}
return M[n-1];
}
};