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

Runtime: 44 ms, faster than 38.51% of Java online submissions for Boats to Save People.

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

LC 881. Boats to Save People的更多相关文章

  1. 【LeetCode】881. Boats to Save People 解题报告(Python)

    [LeetCode]881. Boats to Save People 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu ...

  2. [LeetCode] 881. Boats to Save People 渡人的船

    The i-th person has weight people[i], and each boat can carry a maximum weight of limit. Each boat c ...

  3. 881. Boats to Save People

    The i-th person has weight people[i], and each boat can carry a maximum weight of limit. Each boat c ...

  4. LeetCode 881. Boats to Save People

    原题链接在这里:https://leetcode.com/problems/boats-to-save-people/ 题目: The i-th person has weight people[i] ...

  5. [Leetcode 881]船救人 Boats to Save People 贪心

    [题目] The i-th person has weight people[i], and each boat can carry a maximum weight of limit. Each b ...

  6. [Swift]LeetCode881. 救生艇 | Boats to Save People

    The i-th person has weight people[i], and each boat can carry a maximum weight of limit. Each boat c ...

  7. 【leetcode】885. Boats to Save People

    题目如下: 解题思路:本题可以采用贪心算法,因为每条船最多只能坐两人,所以在选定其中一人的情况下,再选择第二个人使得两人的体重最接近limit.考虑到人的总数最大是50000,而每个人的体重最大是30 ...

  8. 算法与数据结构基础 - 双指针(Two Pointers)

    双指针基础 双指针(Two Pointers)是面对数组.链表结构的一种处理技巧.这里“指针”是泛指,不但包括通常意义上的指针,还包括索引.迭代器等可用于遍历的游标. 同方向指针 设定两个指针.从头往 ...

  9. 算法与数据结构基础 - 贪心(Greedy)

    贪心基础 贪心(Greedy)常用于解决最优问题,以期通过某种策略获得一系列局部最优解.从而求得整体最优解. 贪心从局部最优角度考虑,只适用于具备无后效性的问题,即某个状态以前的过程不影响以后的状态. ...

随机推荐

  1. 【Java并发】线程通信

    一.概述 1.1 什么是多线程之间通讯? 1.2 案例 代码实现 解决线程安全问题 二.等待通知机制 2.1 示例 2.2 wait与sleep区别 三.Lock锁 3.1 概述 3.2 等待/通知机 ...

  2. RobHess的SIFT代码解析步骤四

    平台:win10 x64 +VS 2015专业版 +opencv-2.4.11 + gtk_-bundle_2.24.10_win32 主要参考:1.代码:RobHess的SIFT源码 2.书:王永明 ...

  3. Matlab---读取 .txt文件

    Matlab读取 .txt文件 这里提供两种方法:1,load()函数.2,importdata()函数. ---------------------------------------------- ...

  4. ndk学习之C语言基础复习----基本数据类型、数组

    关于NDK这个分类在N年前就已经创建了,但是一直木有系统的记录其学习过程,当然也没真正学会NDK的技术真谛,所以一直也是自己的一个遗憾,而如今对于Android程序员的要求也是越来越高,对于NDK也是 ...

  5. metal cmd执行时间

    https://developer.apple.com/library/archive/documentation/3DDrawing/Conceptual/MTLBestPracticesGuide ...

  6. sql random string

    begindeclare chars_str varchar(62) default 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123 ...

  7. C语言之volatile

    emOsprey  鱼鹰谈单片机 2月21日 预计阅读时间: 4 分钟 和 const 不同(关于 const 可以看 const 小节),当一个变量声明为 volatile,说明这个变量会被意想不到 ...

  8. Mysql判断是否某个字符串在某字符串字段的4种方法

    方法一:like SELECT * FROM 表名 WHERE 字段名 like "%字符%"; 方法二:find_in_set() 利用mysql 字符串函数 find_in_s ...

  9. BZOJ 3489: A simple rmq problem (KD-tree做法)

    KD树水过这道可持久化树套树-其实就是个三维偏序 题解戳这里 CODE #include <bits/stdc++.h> using namespace std; #define ls ( ...

  10. Codeforces Round #588 (Div. 2) C. Anadi and Domino(思维)

    链接: https://codeforces.com/contest/1230/problem/C 题意: Anadi has a set of dominoes. Every domino has ...