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[LeetCode] Unique Paths II 不同的路径之二

2015-03-20 15:13 513 查看
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.

这道题是之前那道 Unique Paths 不同的路径 的延伸,在路径中加了一些障碍物,还是用动态规划Dynamic Programming来解,不同的是当遇到为1的点,将该位置的dp数组中的值清零,其余和之前那道题并没有什么区别,代码如下:

解法一:

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


或者我们也可以使用一维dp数组来解,省一些空间,参见代码如下:

解法二:

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


类似题目:

Unique Paths

LeetCode All in One 题目讲解汇总(持续更新中...)
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