Leetcode:63. Unique Paths II

Description

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.

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.

思路

同前一个题一样的动态规划，用一个二维数组记录状态,dp[i][j]表示从(0,0)到(i,j)的路径数

dp[i][j] = dp[i][j-1] + dp[i-1][j]，若obstacleGrid[i][j] = 1, dp[i][j] = 0

可以把二维数组降到一维

代码

二维数组

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

vector<vector<int>> dp(m, vector<int>(n, 0));
dp[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
for(int i = 1; i < m; ++i){
if(obstacleGrid[i][0] == 1)
dp[i][0] = 0;
else dp[i][0] = dp[i - 1][0];
}

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

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

return dp[m - 1][n - 1];
}
};

一维数组

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

vector<int> dp(n, 0);
dp[0] = obstacleGrid[0][0] ? 0 : 1;
for(int j = 1; j < n; ++j){
if(obstacleGrid[0][j] == 1)
dp[j] = 0;
else
dp[j] = dp[j - 1];
}

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

sum = dp[j];
}
}

if(n == 1 && m == 1 && obstacleGrid[0][0] == 0) return 1;
return dp[n - 1];
}
};