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

题目如下：

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：首先可以确认Capacity的最小值是1，最大值是50000*500。因为知道上下限，所以这里可以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，那第一天可以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]，那第二天能装载的最大重量就是w[j] + mid,同理可以在w中最大的小于或者等于w[j]+mid的元素，因为w是有序的，所以这里继续使用二分查找，如果D天装载的包裹重量能大于或者等于w[-1]，那么表示mid满足条件，下一步让 = mid - 1；如果不满足说明则令mid = low + 1，直到找出最小的mid为止。

代码如下：

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