1011. Capacity To Ship Packages Within D Days

A conveyor belt has packages that must be shipped from one port to another withinDdays.

Thei-th package on the conveyor belt has a weight ofweights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given byweights). 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 withinDdays.

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. 

Example 2:

Input: weights = [3,2,2,4,1,4], D = 3
Output: 6
Explanation: 
A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
1st day: 3, 2
2nd day: 2, 4
3rd day: 1, 4

Example 3:

Input: weights = [1,2,3,1,1], D = 4
Output: 3
Explanation: 
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1

Note:

1. 1 <= D <= weights.length <= 50000
2. 1 <= weights[i] <= 500

跟上面那道题很像，是真的像

同样是二分法，先确认边界，left应该是weights的最大值，right应该是weights的sum

while（left < right）

medium=1/2（left+right）乃当前capacity

设置local变量day用来计算当前capacity最多用几天，再设置cur用来存放此时的load量

for each w 循环weights

如果need加w大于了medium，说明day要+1，重置cur

cur += w

最后判断一下day和D的关系，大于说明要的天数太多了得减少，通过增加capacity（left = mid + 1）来实现，否则right = mid

return left

classSolution {
  public int shipWithinDays(int[] weights, intD) {
        int left = 0, right = 0;
        for (intw: weights) {
            left =Math.max(left, w);
            right +=w;
        }
        while (left <right) {
            int mid = (left + right) / 2, day = 1, cur = 0;
            for (intw: weights) {
                if (cur + w >mid) {
                    day += 1;
                    cur = 0;
                }
                cur +=w;
            }
            if (day > D) left = mid + 1;
            else right =mid;
        }
        returnleft;
    }
}