【leetcode】1014. Capacity To Ship Packages Within D Days

题目以下：

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.

 

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

解题思路：首先能够确认Capacity的最小值是1，最大值是50000*500。由于知道上下限，因此这里能够采用二分查找的方法。记w[i]为前i个包裹的重量和，显然w是一个递增的数组，假设Capacity为mid，那第一天能够装载的包裹的重量就是在w中最大的小于或者等于mid的元素，这里记为w[j]，那次日能装载的最大重量就是w[j] + mid,同理能够在w中最大的小于或者等于w[j]+mid的元素，由于w是有序的，因此这里继续使用二分查找，若是D天装载的包裹重量能大于或者等于w[-1]，那么表示mid知足条件，下一步让 = mid - 1；若是不知足说明则令mid = low + 1，直到找出最小的mid为止。

代码以下：

class Solution(object): def shipWithinDays(self, weights, D): """ :type weights: List[int] :type D: int :rtype: int """ w = [] low = 0 for i in weights: low = max(low,i) if len(w) == 0: w += [i] else: w += [w[-1] + i] #print w
        high = 50000*500 res = float('inf') import bisect while low <= high: mid = (low+high)/2 start = 0 times = 1 tmp_weight = mid while times <= D: inx = bisect.bisect_left(w,tmp_weight,start) if inx == len(w): break
                elif tmp_weight == w[inx]: if inx == len(w) - 1: break tmp_weight = w[inx] + mid times += 1
                else: tmp_weight = w[inx-1] + mid times += 1
            #print mid,times
            if times <= D and tmp_weight >= w[-1]: res = min(res,mid) high = mid -1
            else: low = mid + 1

        return res