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

Solution

Time complexity : O(nlogn)
Space complexity : O(1)

class Solution {
public:
    int numRescueBoats(vector<int>& people, int limit) {
        sort(people.rbegin(), people.rend());
        int ans = 0;
        for (int i=0, j=people.size()-1; i<=j; ++i) {
            ++ans;
            if (i == j) break;
            if (people[i] + people[j] <= limit)
                --j;
        }
        return ans;
    }
};

將體重由大→小排序，每次由最重的先裝，若還裝的下最輕的，就進去。