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.

思路：DFS，大数据的TLE，需要进行优化

class Solution {
public:
int ans;
void dfs(int i, int j, int step,vector<vector<int> > &obstacleGrid){
if (i == obstacleGrid.size() || j == obstacleGrid[0].size()){
return;
}
if (obstacleGrid[i][j]){
return;
}
if (step == obstacleGrid.size() + obstacleGrid[0].size() -1){
ans++;
return;
}
dfs(i+1,j,step + 1,obstacleGrid);
dfs(i,j+1,step + 1,obstacleGrid);
return;
}
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
ans = 0;
dfs(0,0,1,obstacleGrid);
return ans;
}
};

问题：大数据的时候TLE,问题搜索的时候，走了很多重复的子路径

所以不能进行DFS(DFS本质上是暴力),而应该DP

DP[i,j] = 代表从A[0,0]到A[i,j]的path个数

DP[i,j] = A[i-1,j] is'not obstacle ? DP[i-1,j] + A[i,j-1] is'not obstacle DP[i,j-1]

0 1 0 0 0

1 0 0 0 0

0 0 0 0 0 ==》第一行的时候，需要特殊处理，1之后的初值都是0，1之前的才是1

0 0 0 0 0

class Solution {
public:
int dp(vector<vector<int> > &obstacleGrid){
vector<int> dp(obstacleGrid[0].size(),1);
int m = obstacleGrid.size(),n = obstacleGrid[0].size();
//第一行需要特殊处理
bool first = false;
for(int j = 0; j < n; j++){
if (obstacleGrid[0][j]){
first = true;
}
if (obstacleGrid[0][j] || first){
dp[j] = 0;
}else{
dp[j] = 1;
}
}
for(int i = 1; i < m; i++){
for(int j = 0; j < n; j++){
//dp[i][j] = dp[i-1][j] + dp[i][j-1];
if (obstacleGrid[i][j]){  //是障碍，说明到当前位置的路径个数是0
dp[j] = 0;
}else if (j){
dp[j] = dp[j] + dp[j-1];
}
}
}
return dp[dp.size() -1];
}
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return dp(obstacleGrid);
}
};