HackMD
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 ith 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
