Code
使用 buequet += (flower + 1) / k; 以及 flower = (flower + 1) % k 來計算連續的花束的數量。
要注意 edge case： m * k 有可能會 overflow，int 會裝不下。
class Solution {
public:
int minDays(vector<int>& bloomDay, int m, int k) {
long long total = (long long) m * (long long) k;
if((long long)bloomDay.size() < total) return -1;
int l = 1, r = 1;
for(int d: bloomDay) {
r = max(r, d);
}
while(l < r) {
int mid = l + (r - l) / 2;
int flower = 0;
int buequet = 0;
for(int bloom: bloomDay) {
if(bloom > mid) flower = 0;
else {
buequet += (flower + 1) / k;
flower = (flower + 1) % k;
}
}
if (buequet >= m) r = mid;
else l = mid + 1;
}
return l;
}
};