881.Boats to Save People

tags: `Medium`,`Array`,`Two Pointers`,`Greedy`,`Sorting`

題目描述

You are given an array people where people[i] is the weight of the i^th person, and an infinite number of boats where each boat can carry a maximum weight of limit. Each boat carries at most two 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.

範例

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)

Constraints:

1 <= people.length <= 5 * 10⁴
1 <= people[i] <= limit <= 3 * 10⁴

解答

C++















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

Yen-Chi ChenMon, Apr 3, 2023

Python










class Solution:
    def numRescueBoats(self, people: List[int], limit: int) -> int:
        ans, l, r = 0, 0, len(people)-1
        people.sort()
        while l <= r:
            if people[l] + people[r] <= limit:
                l += 1
            ans += 1
            r -= 1
        return ans

Yen-Chi ChenMon, Apr 3, 2023

Javascript
















function numRescueBoats(people, limit) {
  people.sort((a, b) => b - a);
  let boats = 0;
  let right = people.length - 1;

  for (let left = 0; left <= right; left++) {
    if (people[left] + people[right] <= limit) {
      boats++;
      right--;
    } else {
      boats++;
    }
  }

  return boats;
}

MarsgoatApr 12, 2023

Reference

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