Problem
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 <= 500001 <= people[i] <= limit <= 30000
Solution: Greedy + Two Pointers
Time complexity: O(nlogn)
Space complexity: O(1)
Put one heaviest guy and put the lightest guy if not full.
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// Author: Huahua // Running time: 80 ms class Solution { public: int numRescueBoats(vector<int>& people, int limit) { sort(people.rbegin(), people.rend()); int i = 0; int j = people.size() - 1; int ans = 0; while (i <= j) { ++ans; if (i == j) break; if (people[i++] + people[j] <= limit) --j; } return ans; } }; |
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