The i-th package on the conveyor belt has a weight of weights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.
Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within D days.
Example 1:
Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
Output: 15
Explanation:
A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
1st day: 1, 2, 3, 4, 5
2nd day: 6, 7
3rd day: 8
4th day: 9
5th day: 10
Note that the cargo must be shipped in the order given, so using a ship of capacity 14
and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.
思路：典型的二分答案，capacity越大，需要的days越小，就是一個二分的過程，對days進行計算後，判斷capacity往哪邊移動，下限是max(A[i]), 上限是sum(A[i]); 注意最後一個element，跳出循環了，cursum 還有，count需要+1;
class Solution {
public int shipWithinDays(int[] weights, int D) {
if(weights == null || weights.length == 0) {
return 0;
}
int start = 0; int end = 0;
for(int i = 0; i < weights.length; i++) {
start = Math.max(start,weights[i]);
end += weights[i];
}
while(start + 1 < end) {
int mid = start + (end - start) / 2;
if(calculateDays(mid, weights) > D) {
start = mid;
} else {
//calculateDays(mid, weights) <= D
end = mid;
}
}
if(calculateDays(start, weights) > D) {
return end;
}
return start;
}
private int calculateDays(int capacity, int[] weights) {
int count = 0;
int cursum = 0;
for(int i = 0; i < weights.length; i++) {
cursum += weights[i];
if(cursum <= capacity) {
continue;
} else {
count++;
cursum = weights[i];
}
}
return count + 1;
}
}