[LeetCode] 881. Boats to Save People

[grandyang]

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.)

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

这道题让我们载人过河，说是每个人的体重不同，每条船承重有个限度 limit（限定了这个载重大于等于最重人的体重），同时要求每条船不能超过两人，这尼玛是独木舟吧，也就比 kayak 大一点点吧（不过也有可能是公园湖中的双人脚蹬船，怀念小时候在公园划船的日子～），问我们将所有人载到对岸最少需要多少条船。从题目中的例子2可以看出，最肥的人有可能一人占一条船，当然如果船的载量够大的话，可能还能挤上一个瘦子，那么最瘦的人是最可能挤上去的，所以策略就是胖子加瘦子的上船组合。那么这就是典型的贪婪算法的适用场景啊，首先要给所有人按体重排个序，从瘦子到胖子，这样我们才能快速的知道当前最重和最轻的人。然后使用双指针，left 指向最瘦的人，right 指向最胖的人，当 left 小于等于 right 的时候，进行 while 循环。在循环中，胖子是一定要上船的，所以 right 自减1是肯定有的，但是还是要看能否再带上一个瘦子，能的话 left 自增1。然后结果 res 一定要自增1，因为每次都要用一条船，参见代码如下：

class Solution {
public:
    int numRescueBoats(vector<int>& people, int limit) {
        int res = 0, n = people.size(), left = 0, right = n - 1;
        sort(people.begin(), people.end());
        while (left <= right) {
        	if (people[left] + people[right] <= limit) ++left;
        	--right;
        	++res;
        }
        return res;
    }
};

Github 同步地址:

#881

参考资料：

https://leetcode.com/problems/boats-to-save-people/

https://leetcode.com/problems/boats-to-save-people/discuss/156740/C%2B%2BJavaPython-Two-Pointers

https://leetcode.com/problems/boats-to-save-people/discuss/156855/6-lines-Java-O(nlogn)-code-sorting-%2B-greedy-with-greedy-algorithm-proof.

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