标签：array dynamic programming leetcode
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.
思路：这道题和LeetCode: 56 Unique Paths非常相似，只是这次首先要先判断是否有障碍。不是每次都有两个选择（向右，向下）。因为有了这个，我们不能直接求和得到结果。递推式和unique path一样，不过我们每次都要先判断是否有障碍，如果有，dp[i][j] = 0,如果没有dp[i][j] = dp[i-1][j] + dp[i][j-1]. 所以，实际上我们还是只需要一个一维数组，因为更新时的信息足够了。
Attention：
1. 注意和unique path的区别，首先是一维数组的初始化，由于我们还需要判断第一行是否存在障碍，所以初始化时为0.
<span style="font-size:14px;">vector<int> dp(col, 0);</span>2. 由于我们还需要计算第一行的dp值，所以循环范围从0~row, 0~col.注意计算递归式时，dp[ci] = dp[ci] + dp[ci-1],要先判断ci>0.
<span style="font-size:14px;">for(int ri = 0; ri < row; ri++)
{
for(int ci = 0; ci < col; ci++)
{</span>3. 我们需要初始化dp[0], 如果obstacleGrid[0][0] = 1, 则会重置dp[0] 为0， 否则dp[0]= 1,将用于更新下一个动态规划表的值。
<span style="font-size:14px;">if(obstacleGrid[ri][ci] == 1)
{
dp[ci] = 0;
}
else
{
if(ci > 0)
dp[ci] = dp[ci] + dp[ci-1];
}</span>复杂度：时间复杂度，一次遍历O(n), 空间复杂度，一维数组，O(N)
AC Code:
<span style="font-size:14px;">class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int row = obstacleGrid.size();
int col = obstacleGrid[0].size();
if(row == 0 || col == 0) return 0;
vector<int> dp(col, 0);
dp[0] = 1;
for(int ri = 0; ri < row; ri++)
{
for(int ci = 0; ci < col; ci++)
{
if(obstacleGrid[ri][ci] == 1)
{
dp[ci] = 0;
}
else
{
if(ci > 0)
dp[ci] = dp[ci] + dp[ci-1];
}
}
}
return dp.back();
}
};</span>[C++]LeetCode: 78 Unique Paths II
标签：array dynamic programming leetcode
原文地址：http://blog.csdn.net/cinderella_niu/article/details/42523495
