Topic
answer
BFS problem, pay attention to the case where rows and cols are 0, the others are BFS templates
Code
class Solution {
public:
//求横纵坐标各位数之和
int get_sum(int x, int y) {
int sum = 0;
while (x) {
sum += x % 10;
x /= 10;
}
while (y) {
sum += y % 10;
y /= 10;
}
return sum;
}
int movingCount(int threshold, int rows, int cols) {
if (!rows || !cols) return 0; //边界情况
int dx[4] = {
1, -1, 0, 0};
int dy[4] = {
0, 0, 1, -1};
//初始化点都没有走过
vector<vector<bool>> st(rows, vector<bool>(cols, false));
int res = 0;
queue<pair<int, int>> q;
q.push({
0, 0});
while (q.size()) {
auto t = q.front();
q.pop();
int x = t.first, y = t.second;
//已经走过或者是和大于给定范围
if (st[x][y] || get_sum(x, y) > threshold) continue;
res++;
st[x][y] = true; //标记已经
for (int i = 0; i < 4; i++) {
int a = x + dx[i], b = y + dy[i];
if (a >= 0 && a < rows && b >= 0 && b < cols) {
q.push({
a, b});
}
}
}
return res;
}
};