greedy

881. Boats to Save People Medium

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.

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

为什么可以使用最大匹配最小？
因为一只船最多只能容纳两个人，所以我们可以从最重的船开始放，那么为什么要匹配最轻的船？因为如果最轻的都匹配不了的话

class Solution {
    public int numRescueBoats(int[] people, int limit) {
        Arrays.sort(people);
        int left = 0,right = people.length-1;
        int result = 0;
        while(left<=right){
            if(people[right]+people[left]<=limit){
                left++;right--;
            }
            else{
                right--;
            }
            result++;
        }
        return result;
    }
}

 

标签：people,int,greedy,limit,boats,Input,boat
来源： https://www.cnblogs.com/cynrjy/p/16414943.html