Leetcode 1011. Capacity To Ship Packages Within D Days 在 D 天内送达包裹的能力

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1011. Capacity To Ship Packages Within D Days

题目描述：

A conveyor belt has packages that must be shipped from one port to another within D days.

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.

示例：

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.

解答：

这道题用二分查找法，二分查找的左边是最重包裹的重量，二分查找的右边是所有包裹的重量和。

每次做二分查找的时候，判断查找的重量（也就是船的负载）是否能让船在D天内把货物送达。

如果天数小于D，让左边=查找的重量+1 若天数大于等于D，让右边=查找的重量。

最后得到的重量就是货物的负载量。

代码：

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class Solution {
public:
    int shipWithinDays(vector<int>& weights, int D) {
        int maxWeight = 0;
        int upperBound = 0;
        for (auto &w : weights) {
        	if (w > maxWeight)
        		maxWeight = w;
        	upperBound += w;
        }
        while (maxWeight < upperBound) {
        	
        	int days = 1;
        	int localWeight = 0;
        	int mid = (maxWeight + upperBound) / 2;
        	for (auto &w : weights) {
        		localWeight += w;
        		if (localWeight > mid) {
        			days ++;
        			localWeight = w;
        		}

        	}
        	if (days > D) maxWeight = mid + 1;
        	if (days <= D) upperBound = mid;
        }
        return maxWeight;
    }
};