LeetCode - 62/63/64 - Unique Paths/Minimum Path Sum

62.Unique Paths

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.
一个机器人到右下角摘星星，只能向右向下移动，问有多少条路可达。
一个dp题，每一个格子可从他的上边或者左边走过来，那么到达这个格子的路径数 = 到达他上边格子的路径数 + 到达他左边格子的路径数。时间复杂度O(m*n)，空间复杂度O(m*n)

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

看到一种写法，回来补下题解，上面那种空间复杂度过高，可以降

因为每次只需要用到上一行的值，所以会发现我们只要开两个vector存上一行的值和这一行的值便足够了，不需要开整个二维空间存放。

class Solution {
int uniquePaths(int m, int n) {
if (m > n) return uniquePaths(n, m);
vector<int> pre(m, 1);
vector<int> cur(m, 1);
for (int j = 1; j < n; j++) {
for (int i = 1; i < m; i++)
cur[i] = cur[i - 1] + pre[i];
swap(pre, cur);
}
return pre[m - 1];
}
};

然后发现，我们每次都覆盖了pre的值，就相当于，只用到了cur上一次存的值，这个时候我们可以省下pre数组的空间，只存cur即可。此时时间复杂度O(mn)，空间复杂度(max(n,m))

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

63. Unique Paths II

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.
在上一题的基础上添加了障碍物，思路跟上题一样，就是处理dp的时候，将障碍物存在的点ans[i][j]设为0，其余不变。时间复杂度O(m*n)，空间复杂度O(m*n)

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

新写法，跟上一题第二种做法思路相同，时间复杂度O(mn)，空间复杂度O(n)

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

64. Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.
求一条从左上角到右下角沿途数字和最小的路径。延续了上面两题的新思路，依旧是只开O(n)的空间，来存每一行的结果（因为每一行的结果仅与上一行有关）。时间复杂度O(mn)，空间复杂度O(n)

class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
vector<int> ans(n, 0);
ans[0] = grid[0][0];
for (int i = 1; i < n; ++i) {
ans[i] = ans[i-1] + grid[0][i];
}
for (int i = 1; i < m; ++i) {
ans[0] += grid[i][0];
for (int j = 1; j < n; ++j) {
ans[j] = min(ans[j-1], ans[j]) + grid[i][j];
}
}
return ans[n-1];
}
};