LeetCode Unique Paths II

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

和Unique Paths一样只不过这回不太能用公式直接得出了，在动态规划时依据提供的障碍信息确定如何选取子问题的解，方便起见直接将入参用作dp数组

第二轮：

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.

如果当前位置有障碍那么该处的dp数组元素直接为0，其余和unique path一样

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m = obstacleGrid.size();
if (m < 1) return 0;
int n = obstacleGrid[0].size();
if (n < 1) return 0;

int* dp = new int[n+1];
for (int i=0; i<=n; i++) dp[i] = 0;
if (obstacleGrid[0][0] != 1) dp[1] = 1;

for (int i=0; i<m; i++) {
for (int j=1; j<=n; j++) {
if (obstacleGrid[i][j-1] != 1) {
dp[j] = dp[j] + dp[j-1];
} else {
dp[j] = 0;
}
}
}
return dp
;
}
};