unique paths ii

思路

比上一题加了石头这个概念,其实不难理解,在初始化第一行跟第一列的时候,如果是石头就置为0,在计算其他位置的时候也是一样,如果是石头则为0否则如上一题。

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int (vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int> > result(m);
for (int i = 0; i < m; i++) {
result[i].resize(n, 0);
}
result[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
if (result[0][0] == 0) {
return 0;
}
for (int i = 1; i < n; i++) {
result[0][i] = obstacleGrid[0][i] == 1 ? 0 : result[0][i - 1];
}
for (int i = 1; i < m; i++) {
result[i][0] = obstacleGrid[i][0] == 1 ? 0 : result[i-1][0];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
result[i][j] = obstacleGrid[i][j] == 1 ? 0 : result[i][j - 1] + result[i - 1][j];
}
}
return result[m - 1][n - 1];
}