2018年11月4日

881. Boats to Save People

問題描述：

The i-th person has weight people[i], and each boat can carry a maximum weight of limit.

Each boat carries at most 2 people at the same time, provided the sum of the weight of those people is at most limit.

Return the minimum number of boats to carry every given person. (It is guaranteed each person can be carried by a boat.)

樣例：

Example 1:

Input: people = [1,2], limit = 3
Output: 1
Explanation: 1 boat (1, 2)

Example 2:

Input: people = [3,2,2,1], limit = 3
Output: 3
Explanation: 3 boats (1, 2), (2) and (3)

Example 3:

Input: people = [3,5,3,4], limit = 5
Output: 4
Explanation: 4 boats (3), (3), (4), (5)

Note:

• 1 <= people.length <= 50000
• 1 <= people[i] <= limit <= 30000

問題分析：

本題難度為Medium！屬於貪心問題，已給出的函式定義為：

class Solution:
    def numRescueBoats(self, people, limit):
        """
        :type people: List[int]
        :type limit: int
        :rtype: int
        """

此題的需求是使船的數量最小，如果最重的人可以與最輕的人共用一艘船，那麼就安排他們坐同一條船。否則，最重的人無法與任何人配對（因為其他人都比最輕的人重），那麼他將自己獨自乘一艘船。

演算法; 首先對people進行從小到大排序。 令 people[light_index] 指向當前最輕的人（即列表前向），而 people[heavy_index] 指向最重的那人（即列表反向）。 如果最重的人可以與最輕的人可以坐同一條船（即 people[light_index] + people[heavy_index] <= limit），則船數量加一，更新最重的人和最輕的人；否則，最重的人自己獨自坐一條船，更新最重的人。最輕的人待定。直到所有人離開。

程式碼實現：

使用雙指標

# coding=utf-8
class Solution:
    def numRescueBoats(self, people, limit):
        """
        :type people: List[int]
        :type limit: int
        :rtype: int
        """
        people.sort()
        light_index, heavy_index = 0, len(people) - 1
        num_boat = 0
        while light_index <= heavy_index:
            num_boat += 1
            if people[light_index] + people[heavy_index] <= limit:
                light_index += 1
            heavy_index -= 1
        return num_boat