[LeetCode] 881. Boats to Save People
grandyang opened this issue · comments
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:
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